Vector Algebra · Mathematics · TS EAMCET

Two adjacent sides of a triangle are represented by the vectors $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $2 \sqrt{3} \hat{\mathbf{i}}-2 \sqrt{3} \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}$. Then, the least angle of the triangle and perimeter of the triangle are respectively.

A plane $\pi_1$ contains the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$. Another plane $\pi_2$ contains the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $3 \hat{\mathbf{i}}+2 \hat{\mathbf{k}}$. $\mathbf{a}$ is a vectors parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\mathbf{a}$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is acute, then $\theta=$

In a quadrilateral $A B C D, \mathbf{A}=\frac{2 \pi}{3}$ and $A C$ is the bisector of angle $\mathbf{A}$. If $15|\mathbf{A C}|=5|\mathbf{A D}|=3|\mathbf{A B}|$, then angle between $\mathbf{A B}$ and $\mathbf{B C}$ is

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non- coplanar and mutually perpendicular vectors of same magnitude $K . r$ is any vectors satisfying $\mathbf{a} \times((\mathbf{r}-\mathbf{b}) \times \mathbf{a})+\mathbf{b} \times((\mathbf{r}-\mathbf{c}) \times \mathbf{b})+\mathbf{c} \times((\mathbf{r}-\mathbf{a}) \times \mathbf{c})=\mathbf{0}$, then $\mathbf{r}=$

Consider the following

Assertion (A) The two lines $\mathbf{r}=\mathbf{a}+t(\mathbf{b})$ and $\mathbf{r}=\mathbf{b}+s(\mathbf{a})$ intersect each other.

Reason (R) The shortest distance between the lines $\mathbf{r}=\mathbf{p}+t(\mathbf{q})$ and $\mathbf{r}=\mathbf{c}+s(\mathbf{d})$ is equal to the length of projection of the vector ( $\mathbf{p}-\mathbf{c}$ ) on ( $\mathbf{q} \times \mathbf{d}$ )

The correct answer is

$A B C D$ is a tetrahedron, $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$, $3 \bar{i}+2 \bar{j}-\bar{k}$ are the the position vectors of the points $A, B$ and $C$ respectively. $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ is the position vector of the centroid of the triangular face $B C D$. If G is the centroid of the tetrahedron, then $G D=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{c}=-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}$ are three vectors, then $\sqrt{(|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|)+|\mathbf{a}+\mathbf{b}+\mathbf{c}|}=$

Let $\mathbf{a}$ and $\mathbf{b}$ be two vectors such that $|\mathbf{a}|=|\mathbf{b}|$ and $|\mathbf{a}+2 \mathbf{b}|=|2 \mathbf{a}-\mathbf{b}|$. If $\mathbf{c}$ is a vector parallel to $\mathbf{a}$, then the angle between $\mathbf{b}$ and $\mathbf{c}$ is

If $\mathbf{a}$ and $\mathbf{b}$ are two vectors such that $|\mathbf{a}|=|\mathbf{b}|=\sqrt{6}$ and $\mathbf{a} \cdot \mathbf{b}=-1$, then $|\mathbf{a} \times \mathbf{b}| \sin (\mathbf{a}, \mathbf{b})=$

If the volume of a tetrahedron having $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+p \hat{\mathbf{k}}$ as its coterminous edges is 2 , then the values of $\mathbf{p}$ are the roots of the equation

In a $\triangle A B C$, if $\mathbf{B C}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{C A}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, then the perimeter of the triangle is

$\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+c_2 \hat{\mathbf{k}}, a_3 \hat{\mathbf{i}}+b_3 \hat{\mathbf{j}}+c_3 \hat{\mathbf{k}}$ are the position vectors of the points $A, B, C, D$ respectively. $\frac{2}{3}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$ is the position vector of the centroid of the triangular face $B C D$ of the tetrahedron $A B C D$. If $\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}}$ is the position vector of the centroid of the tetrahedron, then $2 \alpha+\beta+\gamma=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=9 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-18 \hat{\mathbf{k}}$ are two vectors, then $\frac{\text { Projection of } \mathbf{b} \text { on } \mathbf{a}}{\text { Projection of } \mathbf{a} \text { on } \mathbf{b}}=$

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \cdot \mathbf{a}=0$, $\mathbf{r} \cdot \mathbf{b}=-2$ and $\mathbf{r} \cdot \mathbf{c}=6$, then $\mathbf{r} \cdot(\beta \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=$

Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{c}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{d}$ is a vector perpendicular to both $\mathbf{a}, \mathbf{b}$ and $|\mathbf{d} \times \mathbf{c}|=14$, then $|\mathbf{d} \cdot \mathbf{c}|=$

If $\mathbf{a}=(x+2 y-3) \hat{\mathbf{i}}+(2 x-y+3) \hat{\mathbf{j}}$ and $\mathbf{b}=(3 x-2 y) \hat{\mathbf{i}} +(x-y+1) \hat{\mathbf{j}}$ are two vectors such that $\mathbf{a}=2 \mathbf{b}$, then $y-5 x=$

$7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+10 \hat{\mathbf{k}},-\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are the position vectors of the points $A, B, C$ and $D$ respectively. If $p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $A B C D$, then $p+q+r=$

If $\mathbf{a}=\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+\sqrt{11} \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$ are two vectors, then the component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+p \hat{\mathbf{k}}$ be two vectors.

If $(\mathbf{a}, \mathbf{b})=60^{\circ}$, then $p=$

$A, B, C$ and $D$, are any four points. If $E$ and $F$ are mid-points of $A C$ and $B D$ respectively, then $\mathbf{A B}+\mathbf{C B}+\mathbf{C D}+\mathbf{A D}=$

The four points whose position vectors are given by $2 a+3 b-c, a-2 b+3 c, 3 a+4 b-2 c$ and $a-6 b+6 c$ are

If $a=|\mathbf{a}| ; b=|\mathbf{b}|$, then $\left(\frac{\mathbf{a}}{a^2}-\frac{\mathbf{b}}{b^2}\right)^2$

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three unit vectors such that $x \mathbf{a}+y \mathbf{b}+z \mathbf{c}= p(\mathbf{b} \times \mathbf{c})+q(\mathbf{c} \times \mathbf{a})+r(\mathbf{a} \times \mathbf{b})$. If $(\mathbf{a}, \mathbf{b})=(\mathbf{b}, \mathbf{c})=(\mathbf{c}, \mathbf{a})=\frac{\pi}{3}$, $(\mathbf{a}, \mathbf{b} \times \mathbf{c})=\frac{\pi}{6}$ and $\mathbf{a}, \mathbf{b}, \mathbf{c}$ form a right-handed system, then $\frac{x+y+z}{p+q+r}=$

$O(0,0,0), A(3,1,4), B(1,3,2)$ and $C(0,4,-2)$ are the vertices of a tetrahedron. If $G$ is the centroid of the tetrahedron and $G_1$ is the centroid of its face $A B C$, then the point which divides $G G_1$ in the ratio $1: 2$ is

The position vectors of two points $A$ and $B$ are $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ respectively. The point $P$ with position vector $-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is on the line $A B$. If the point $Q$ is the harmonic conjugate of $P$, then the sum of the scalar components of the position vector of $Q$ is

If $\mathbf{a}$ and $\mathbf{b}$ are two vectors such that $|\mathbf{a}|=5,|\mathbf{b}|=12$ and $|\mathbf{a}-\mathbf{b}|=13$, then $|2 \mathbf{a}+\mathbf{b}|=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors, then $(\mathbf{a}+2 \mathbf{b}) \times(3 \mathbf{a}-\mathbf{b})$

$\mathbf{a}$ is a vector perpendicular to the plane containing non zero vectors $\mathbf{b}$ and $\mathbf{c}$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are such that

$|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+|\mathbf{c}|^{2}}$, then

$|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}|+|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors such that $\mathbf{a}$ is perpendicular to both $\mathbf{b}, \mathbf{c}$ and angle between $\mathbf{b}, \mathbf{c}$ is $2 \pi / 3$, then $|a+3 b-4 c|^2=$

Let $\mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ be the position vector of a point $A$. Let $\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be two vectors and $\mathbf{r}$ be a vector passing through the point $A(\mathbf{a})$ and parallel to the vector $\mathbf{b}$. If the projection of $\mathbf{r}$ on $\mathbf{c}$ is $\frac{9}{\sqrt{6}}$, then $|\mathbf{r}|=$

If $S$ is the circumcentre, $O$ is the orthocentre and $G$ is the centroid of a $\triangle A B C$, then match the items of the List-I with those of the items of List-II given below.

<mi data-mjx-auto-op="false">SA</mi>

<mi data-mjx-auto-op="false">SB</mi>

<mi data-mjx-auto-op="false">SC</mi>

<mi data-mjx-auto-op="false">SA</mi>

<mi data-mjx-auto-op="false">SB</mi>

<mi data-mjx-auto-op="false">SC</mi>

<mi data-mjx-auto-op="false">GA</mi>

<mi data-mjx-auto-op="false">GB</mi>

<mi data-mjx-auto-op="false">GC</mi>

<mi data-mjx-auto-op="false">GA</mi>

<mi data-mjx-auto-op="false">GB</mi>

<mi data-mjx-auto-op="false">GC</mi>

<mo>/</mo>

<mi data-mjx-auto-op="false">OS</mi>

<mo>/</mo>

<mi data-mjx-auto-op="false">OS</mi>

<mi data-mjx-auto-op="false">OA</mi>

<mi data-mjx-auto-op="false">OB</mi>

<mi data-mjx-auto-op="false">OC</mi>

<mi data-mjx-auto-op="false">OA</mi>

<mi data-mjx-auto-op="false">OB</mi>

<mi data-mjx-auto-op="false">OC</mi>

Then, the correct match is

Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be three vectors such that $\mathbf{a} \cdot \mathbf{a}=\mathbf{b} \cdot \mathbf{b}=\mathbf{c} \cdot \mathbf{c}=5$ and $|\mathbf{a}+\mathbf{b}-\mathbf{c}|^2+|\mathbf{b}+\mathbf{c}-\mathbf{a}|^2+|\mathbf{c}+\mathbf{a}-\mathbf{b}|^2=50$, then $\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=$

Let $\mathbf{c}$ be a vector coplanar with the unit vectors $\mathbf{a}, \mathbf{b}$ and let $\mathbf{d}$ be the unit vector perpendicular to $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$. If $[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $30^{\circ}$, then $|\mathbf{c}|=$

If $|\mathbf{a}|=4,|\mathbf{b}|=5$ and $|\mathbf{a}-\mathbf{b}|=3$ and $\theta$ is the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$, then $\cot ^2 \theta=$

If $\mathbf{a}+\mathbf{b}+\mathbf{c}=0,|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is

If $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}},-12 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}},-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are the position vectors of four coplanar points, then $\lambda=$

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be two vectors. If the orthogonal projection vector of $\mathbf{a}$ on $\mathbf{b}$ is $\mathbf{x}$ and orthogonal projection vector of $\mathbf{b}$ on $\mathbf{a}$ is $\mathbf{y}$, then $|\mathbf{x}-\mathbf{y}|=$

II. If the points with position vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-\hat{\mathbf{k}}$, $\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are coplanar, then the magnitude of the vector $6 \lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ is

Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be three non-coplanar vectors and $L$ be the line passing through the points $\mathbf{a}-\mathbf{b}+\mathbf{c}$ and $\mathbf{b}-\mathbf{c}$. If $\pi$ is a plane passing through the points $2 \mathbf{a}-\mathbf{b}, 2 \mathbf{b}-\mathbf{c}$ and $2 c-\mathbf{a}$, then the point of intersection of $L$ and $\pi$ is

Let $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}$ be three vectors. If $\mathbf{p}$ is the projection of $\mathbf{b}$ on $\mathbf{a}$ and $\mathbf{q}$ is the projection of $\mathbf{c}$ on $\mathbf{a}$, then $13 \mathbf{p}=$

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. Let $\mathbf{r}$ be a vector perpendicular to both $\mathbf{b}$, $c$ and $\mathbf{r} \cdot \mathbf{a}=11$. Then, the vector among the following that is perpendicular to $\mathbf{r}$ is

The volume of the tetrahedron with $\hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}}$, $\lambda \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ as coterminous edges is 2 . If $\lambda$ is an integer, then $|\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}+3 \hat{\mathbf{k}}|=$

Let $\mathbf{O A}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{O B}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{O C}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ be the position vectors of three points, $A, B$ and $C$. Let $P$ be the point which divides $A B$ in the ratio $2: 1$. If $l, m, n$ are the direction cosines of the vector $\mathbf{P C}$, then $l+3 m+2 n=$

If the vectors $\mathbf{B C}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{C D}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ represent two adjacent sides of a parallelogram ABCD and $\theta$ is the angle between its diagonals $\mathbf{A C}$ and $\mathbf{B D}$, then $\tan \theta=$

Let $\mathbf{a}=\lambda \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ and $\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ be three vectors for some integer $\lambda$. If the volume of the parallelopiped with $\mathbf{a}, \mathbf{b}, \mathbf{c}$ as coterminous edges is 61 cubic units, then the number of possible values of $\lambda$ is

If two vectors $\mathbf{a}$ and $\mathbf{b}$ which are perpendicular to each other are such that $|\mathbf{a}|=8$ and $|\mathbf{b}|=3$, then $|\mathbf{a}-2 b|=$

Let $\mathbf{a}$ and $\mathbf{b}$ be non-collinear vectors. If the vectors $(\lambda-1) \mathbf{a}+2 \mathbf{b}$ and $3 \mathbf{a}+\lambda \mathbf{b}$ are collinear, then the set of all possible values of $\lambda$ is

Vectors $\mathbf{p}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \mathbf{q}=d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{r}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ forming a $\triangle A B C$ are such that $\mathbf{p}=\mathbf{q}+\mathbf{r}$. If the area of $\triangle A B C$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

$\mathbf{b}$ and $\mathbf{c}$ are non-collinear vectors and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$. If $(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}$ $=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$, then $\sin (\alpha+\beta)=$

Let $A B C$ be a triangle and $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be the position vectors of $A, B$ and $C$, respectively. Let $D$ divides $B C$ in the ratio $3: 1$ internally and $E$ divides $A D$ in the ratio $4: 1$ internally. Let $B E$ meet $A C$ in $F$. If $E$ divides $B F$ in the ratio $3: 2$ internally, then the position vector of $F$ is

If $\alpha, \beta$ and $\gamma$ are real numebrs such that

$$ \begin{aligned} & \left(\frac{7}{3}+\beta\right) \hat{\mathbf{i}}-\hat{\mathbf{j}}+(\alpha+\gamma) \hat{\mathbf{k}} \\ & =\frac{5}{3}(\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\beta(2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}+\gamma \hat{\mathbf{j}}+3 \hat{\mathbf{k}}), \text { then } \\ & 5 \alpha-9 \beta+13 \gamma= \end{aligned} $$

If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{x}=\left(\frac{\mathbf{a b}}{|\mathbf{b}|^2}\right) \mathbf{b}, \mathbf{y}=\left(\frac{\mathbf{a b}}{|\mathbf{a}|^2}\right) \mathbf{a}$ and $\theta$ is angle between $\mathbf{a}$ and $\mathbf{b}$, then $x^2+y^2=$

Three non-coplanar vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are the coterminous edges of a parallelopiped. If $\mathbf{a}$ and $\mathbf{b}$ determine the base of the parallelopiped, then its height is

Let $A B C$ be a triangle and $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be the position vectors of $A, B$ and $C$ respectively. If $D$ divides $B C$ in the ratio $2: 3$ internally and $E$ divides $C A$ in the ratio $2: 1$ internally, then the position vector of the point $P$ which divides $D E$ in the ratio $3: 5$ internally is

If $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are the position vectors of the vertices $A, B, C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is

Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors satisfying $|\mathbf{a}-\mathbf{b}|^2+|\mathbf{a}-\mathbf{c}|^2=10$. Then,

Statement (I): $|\mathbf{a}+2 \mathbf{b}|^2+|2 \mathbf{a}+\mathbf{c}|^2=2$

Statement (II) : $|2 a+3 b|^2+|3 a+2 c|^2=10$

Which of the above statements is (are) true?

If $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ are the position vectors of the points $\mathbf{A}$ and $\mathbf{B}$ respectively, $\mathbf{C}$ divides $\mathbf{A B}$ in the ratio $2: 3$ and $\mathbf{M}$ is the mid-point of $A B$, then 5 (position vector of $\mathbf{C})-2($ position vector of $\mathbf{M})=$

30. If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are the non-coplanar vectors and $\mathbf{a}-2 \mathbf{b}+3 \mathbf{c},-4 \mathbf{a}+5 \mathbf{b}-6 \mathbf{c}, x \mathbf{a}-9 \mathbf{b}+z \mathbf{c}$ are collinear points, then $2 x-z=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, then the component of $\mathbf{b}$ perpendicular to $\mathbf{a}$ is

If $2 \mathbf{i}+3 \mathbf{j}-4 \mathbf{k}$ and $-\mathbf{i}+2 \mathbf{j}+\mathbf{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is

Let the vectors $\mathbf{A B}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{A C}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ be two sides of a $\triangle A B C$. If $G$ is the centroid of $\triangle A B C$, then $\frac{27}{7}|\mathbf{A G}|^2+5=$

If $(\alpha, \beta, \gamma)$ is a triad of real numbers satisfying $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}=\alpha(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\beta(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\gamma(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$, then $\alpha^2-\beta^2+\gamma^2=$

If $\theta$ is the angle between the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $a \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+b \hat{\mathbf{k}}$ and $\cos \theta=\frac{2}{3}$, then $2(a+b+3)=$

Let the volume of the tetrahedron with vertices $\hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}},-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+a \hat{\mathbf{k}}$ be $\frac{20}{3}$. Then the integral value of $a$ is

If $3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, 7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}, \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}$ are position vectors of the points $A, B, C$ and $D$ respectively, then the angle between $\mathbf{A B}$ and $\mathbf{C D}$ is

If $A(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}), B(\lambda \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}), C(-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ and $D(-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ are four points in space such that $\mathbf{A B}=x \mathbf{A C}+y \mathbf{A D}$ for some real number $x \neq 0, y \neq 0$, then $17(\lambda+9)=$

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}$ be two vectors such that $\mathbf{a} \cdot \mathbf{b}=1$, $\cos (\mathbf{a} \cdot \mathbf{b})=\frac{1}{3}$ and the components of $\mathbf{b}$ w.r.t. $(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}})$ be integers. Then, the number of possible vectors that represent $\mathbf{b}$ is

If $\mathbf{a}$ and $\mathbf{b}$ are two vectors such that $\mathbf{a}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$, $|\mathbf{b}|=7, \mathbf{a} \cdot \mathbf{b}=4$ and $|\mathbf{a} \times \mathbf{b}|=5 \sqrt{17}$, then $p=$

In a $\triangle A B C, D$ and $E$ divide the sides $B C$ and $C A$ in the ratio $2: 1$ respectively. If $P$ is the point of intersection of $A D$ and $B E$, then the ratio in which $P$ divides $A D$ is

If the points with position vectors $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ and $a \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ are coplanar, then $a=$

Let $\mathbf{a}$ be a vector in the plane containing vectors $\mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{a}$ is perpendicular to $\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and its projection on $\mathbf{b}$ is $3 \sqrt{6}$, then $|\mathbf{a}|^2=$

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, $\mathbf{d}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ be four vectors and let $l=\mathbf{b} \cdot \mathbf{c}$ and $m=\mathbf{c} \cdot \mathbf{a}$. Then, $[m \mathbf{b}+l \mathbf{a} \mathbf{b d}]=$

If $\mathbf{a , b , c}$ are three independent vectors and there exists a non zero scalar traid $(l, m, n)$ such that $l(3 \mathbf{a}+2 \mathbf{b}+\mathbf{c})+m(2 \mathbf{a}+2 \mathbf{b}+3 \mathbf{c})+n(\mathbf{a}+2 \mathbf{b}+5 \mathbf{c})=\mathbf{0}$, then

If $\mathbf{a}$ and $\mathbf{b}$ represent two non collinear vectors, the equation $\mathbf{r}=t \mathbf{a}+(1-t) \mathbf{b}$ represents

Let $\mathbf{a , b , c}$ be three vectors such that the magnitude of $\mathbf{b}$ is twice that of $\mathbf{a}$ and magnitude of $\mathbf{c}$ is three times that of $\mathbf{a}$. If the angle between each pair of vectors is $\frac{\pi}{3}$ and $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=5$, then $|\mathbf{c}|+|\mathbf{a}|+|\mathbf{b}|=$

If $\mathbf{a , b , c}$ are three mutually perpendicular vectors such that the magnitudes of $\mathbf{b}$ and $\mathbf{c}$ are $1 / 2$ times and $\sqrt{3} / 2$ times that of $\mathbf{a}$, respectively, then the angle between the vectors $\mathbf{a}+\mathbf{b}+\mathbf{c}$ and $\mathbf{b}$ is

The locus of the point $P(\mathbf{r})$ which encloses a triangle $A B P$ of area 1 sq. unit with the fixed points $A(\hat{\mathbf{i}})$ and $B(\hat{\mathbf{j}})$ is

Let $A B C D$ be a parallelogram and $E$ be the mid-point of $A B$. If $P$ is the point of intersection of $D E$ and $A C$, then $\frac{D P}{P E}+\frac{A P}{P C}=$

A vector $\mathbf{a}$ has components $2 p$ and 1 with respect to a two dimensional rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise direction. If $\mathbf{a}$ has components $p+1$ and 1 with respect to the new system, then

If $\mathbf{a}=2 \mathbf{u}+3 \mathbf{v}+7 \mathbf{w}, b=\mathbf{u}+\mathbf{v}-2 \mathbf{w}$ and $\mathbf{c}=-\mathbf{u}-2 \mathbf{v}-3 \mathbf{w}$ then $\left|\frac{[\mathbf{u} \mathbf{v} \mathbf{w}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}\right|(\mathbf{a}+\mathbf{b}+\mathbf{c})=$

Let $\mathbf{V}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{W}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$. If $\mathbf{U}$ is a unit vector, then the maximum value of $[\mathbf{U} \mathbf{V} \mathbf{W}]$ is

The equation of the plane in normal form passing through the point $A(\bar{a})$, parallel to a vector $\bar{b}$ and containing a vector $\bar{c}$ is

Let $\mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{c}$ is a vector such that $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the position vectors of the points $A, B, C$ respectively, then match the items of list-I with those of list-II.

$$ \text { The correct match is } $$

Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. If the orthogonal projection vector of $\mathbf{a}$ on $\mathbf{b}$ be $\mathbf{x}$ and the orthogonal projection vector of $\mathbf{b}$ on $\mathbf{a}$ be $\mathbf{y}$, then $|\mathbf{x}-\mathbf{y}|=$

Let $\mathbf{p}, \mathbf{q}, \mathbf{r}$ be three non-coplanar vectors and $\left[\begin{array}{lll}\mathbf{p} & \mathbf{q} & \mathbf{r}] \mathbf{a}=\mathbf{q} \times \mathbf{r},\left[\begin{array}{lll}\mathbf{p} & \mathbf{q} & \mathbf{r}] \mathbf{b}=\mathbf{r} \times \mathbf{p},[ \end{array}[\mathbf{p}\right. \\ \mathbf{q} & \mathbf{r}] \mathbf{c}=\mathbf{p} \times \mathbf{q} \text {. If }\end{array}\right. \mathbf{a}, \mathbf{b}, \mathbf{c}$ denote the coterminous edges of a parallelopiped, then its height with the base having a and $\mathbf{c}$ is

If $\mathbf{b}, \mathbf{c}$ are non collinear vectors, $|\mathbf{c}| \neq 0$, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$ and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$, then the scalars $\alpha$ and $\beta$ are

If $12 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}-18 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-24 \hat{\mathbf{k}}$ be the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle A B C$, then the position vector of the incentre of $\triangle A B C$ is

For non-coplanar vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$, if the point of intersection of the line $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{c})$ and the plane $\mathbf{r}=\mathbf{b}+\mathbf{c}+x(\mathbf{a}-\mathbf{b})+y(\mathbf{c}+\mathbf{a})$ is $l \mathbf{a}+m \mathbf{b}+n \mathbf{c}$, then $3 l+4 m+2 n=$

If the orthocentre of the triangle whose vertices are $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then

If the vectors $\mathbf{A B}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}, \mathbf{A C}=s \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$, $\mathbf{C B}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ from $\triangle A B C$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle A B C$ is $5 \sqrt{6}$ are

Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors such that $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is not parallel to $\mathbf{c}$. If $\alpha$ and $\beta$ are the angles between $\mathbf{a}, \mathbf{b}$ and $\mathbf{a}, \mathbf{c}$ respectively then $\alpha-\beta=$

Let $\mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b}$ be two non collinear vectors,

$\mathbf{O P}=x_1 \mathbf{a}+y_1 \mathbf{b}, \mathbf{O Q}=x_2 \mathbf{a}+y_2 \mathbf{b}$ and $\mathbf{A}^{\prime} \mathbf{O}=\mathbf{O A}$,

$\mathbf{B}^{\prime} \mathbf{O}=\mathbf{O B}$. If $x_1=\frac{-3}{4}, x_2=\frac{1}{3}, y_1=\frac{7}{4}, y_2=\frac{5}{3}$, then

In a quadrilateral $A B C D$, the point $P$ divides $D C$ in the ratio $1: 3$ internally and $Q$ is the mid-point of $A C$. If $\mathbf{A B}+\mathbf{A D}+\mathbf{B C}-2 \mathbf{D C}=\lambda \mathbf{P Q}$, then the value of $\lambda$ is

$\mathbf{p}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. If the vectors $\mathbf{a}$ and $\mathbf{b}$ are the orthogonal projections of $\mathbf{p}$ on $\mathbf{q}$ and $\mathbf{q}$ on $\mathbf{p}$ respectively, then $\frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a} \cdot \mathbf{b}}=$

Let $\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{b}=7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. The vector $\mathbf{x}$ such that $\mathbf{x} \cdot \mathbf{c}=60$ and perpendicular to both $\mathbf{a}, \mathbf{b}$ is