1
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-0.75
Let $$\overrightarrow a = \widehat i + 2\widehat j + \widehat k,\,\overrightarrow b = \widehat i - \widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i + \widehat j - \widehat k.$$ A vector in the plane of $$\overrightarrow a $$ and $$\overrightarrow b $$ whose projection on $$\overrightarrow c $$ is $${1 \over {\sqrt 3 }},$$ is
A
$$4\widehat i - \widehat j + 4\widehat k$$
B
$$3\widehat i + \widehat j - 3\widehat k$$
C
$$2\widehat i + \widehat j - 2\widehat k$$
D
$$4\widehat i + \widehat j - 4\widehat k$$
2
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-0
(i) Two rays in the first quadrant $x+y=|a|$ and $a x-y=1$ Intersects each other in the interval $a \in\left(a_0, \infty\right)$, the value of $a_0$ is (A) 2
(ii) Point $(\alpha, \beta, \gamma)$ lies on the plane $x+y+z=2$.
Let $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \hat{k} \times(\hat{k} \times \vec{a})=0$, then $\gamma=$
(B) 4/3
(iii) $$
\left|\int_0^1\left(1-y^2\right) d y\right|+\left|\int_1^0\left(y^2-1\right) d y\right|
$$
(C) $$
\left|\int_0^1 \sqrt{1-x} d x\right|+\left|\int_1^0 \sqrt{1+x} d x\right|
$$
(iv) If $\sin A \sin B \sin C+\cos A \cos B=1$, then the value of $\sin C=$ (D) 1
A

$$ \begin{aligned} & \text { (i)-(D); (ii)-(B); (iii)-(B),(C); } \text { (iv)-(A) } \end{aligned} $$

B

$$ \begin{aligned} & \text { (i)-(D); (ii)-(A); (iii)-(B); } \text { (iv)-(D) } \end{aligned} $$

C

$$ \begin{aligned} & \text { (i)-(A); (ii)-(D); (iii)-(B),(C); } \text { (iv)-(D) } \end{aligned} $$

D

$$ \begin{aligned} & \text { (i)-(D); (ii)-(A); (iii)-(B),(C); } \text { (iv)-(D) } \end{aligned} $$

3
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+4
-1
If $$\overrightarrow a \,,\,\overrightarrow b ,\overrightarrow c $$ are three non-zero, non-coplanar vectors and
$$\overrightarrow {{b_1}} = \overrightarrow b - {{\overrightarrow b .\,\overrightarrow a } \over {{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a ,\overrightarrow {{b_2}} = \overrightarrow b + {{\overrightarrow b .\,\overrightarrow a } \over {{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a ,$$
$$\overrightarrow {{c_1}} = \overrightarrow c - {{\overrightarrow c .\,\overrightarrow a } \over {{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a + {{\overrightarrow b .\,\overrightarrow c } \over {{{\left| c \right|}^2}}}{\overrightarrow b _1},\,\,\overrightarrow {{c_2}} = \overrightarrow c - {{\overrightarrow c .\,\overrightarrow a } \over {{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a - {{\overrightarrow b \,.\,\overrightarrow c } \over {{{\left| {{{\overrightarrow b }_1}} \right|}^2}}}{\overrightarrow b _1},$$
$$\overrightarrow {{c_3}} = \overrightarrow c - {{\overrightarrow c .\,\overrightarrow a } \over {{{\left| {\overrightarrow c } \right|}^2}}}\overrightarrow a + {{\overrightarrow b .\,\overrightarrow c } \over {{{\left| c \right|}^2}}}{\overrightarrow b _1},\,\,\overrightarrow {{c_4}} = \overrightarrow c - {{\overrightarrow c .\,\overrightarrow a } \over {{{\left| {\overrightarrow c } \right|}^2}}}\overrightarrow a - {{\overrightarrow b \,.\,\overrightarrow c } \over {{{\left| {{{\overrightarrow b }_1}} \right|}^2}}}{\overrightarrow b _1},$$
then the set of orthogonal vectors is
A
$$\left( {\overrightarrow a ,\overrightarrow {{b_1}} ,\overrightarrow {{c_3}} } \right)$$
B
$$\left( {\overrightarrow a ,\overrightarrow {{b_1}} ,\overrightarrow {{c_2}} } \right)$$
C
$$\left( {\overrightarrow a ,\overrightarrow {{b_1}} ,\overrightarrow {{c_1}} } \right)$$
D
$$\left( {\overrightarrow a ,\overrightarrow {{b_2}} ,\overrightarrow {{c_2}} } \right)$$
4
IIT-JEE 2005 Mains
MCQ (Single Correct Answer)
+3
-1

Incident ray is along the unit vector $$\hat{v}$$ and the reflected ray is along the unit vector $$\widehat{w}$$. The normal is along unit vector $$\hat{a}$$ outwards. Express $$\hat{w}$$, in terms of $$\hat{a}$$ and $$\hat{v}$$.

A
$$\widehat{w}=\hat{v}-2(\hat{a} \cdot \hat{v}) \cdot \hat{a}$$
B
$$\widehat{w}=\hat{v}+2(\hat{a} \cdot \hat{v}) \cdot \hat{a}$$
C
$$\widehat{w}=\hat{v}-3(\hat{a} \cdot \hat{v}) \cdot \hat{a}$$
D
$$\widehat{w}=5\hat{v}+3(\hat{a} \cdot \hat{v}) \cdot \hat{a}$$

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