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1

### JEE Advanced 2021 Paper 2 Online

MCQ (More than One Correct Answer)
Let O be the origin and $$\overrightarrow {OA} = 2\widehat i + 2\widehat j + \widehat k$$ and $$\overrightarrow {OB} = \widehat i - 2\widehat j + 2\widehat k$$ and $$\overrightarrow {OC} = {1 \over 2}\left( {\overrightarrow {OB} - \lambda \overrightarrow {OA} } \right)$$ for some $$\lambda$$ > 0. If $$\left| {\overrightarrow {OB} \times \overrightarrow {OC} } \right| = {9 \over 2}$$, then which of the following statements is (are) TRUE?
A
Projection of $$\overrightarrow {OC}$$ on $$\overrightarrow {OA}$$ is $$- {3 \over 2}$$
B
Area of the triangle OAB is $${9 \over 2}$$
C
Area of the triangle ABC is $${9 \over 2}$$
D
The acute angle between the diagonals of the parallelogram with adjacent sides $${\overrightarrow {OA} }$$ and $${\overrightarrow {OC} }$$ is $${\pi \over 3}$$

## Explanation

Given,

$$\overrightarrow {OA} = 2\widehat i + 2\widehat j + 2\widehat k$$

$$\overrightarrow {OB} = \widehat i - \widehat j + 2\widehat k$$

and $$\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$$

Also, $$\left| {OB \times OC} \right| = 9/2$$ ..... (i)

Now, $$\overrightarrow {OB} \times \overrightarrow {OC} = \overrightarrow {OB} \times {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$$

$$= {{ - \lambda } \over 2}\overrightarrow {OB} \times \overrightarrow {OA} = {\lambda \over 2}(\overrightarrow {OA} \times \overrightarrow {OB} )$$

We need to find $$\overrightarrow {OA} \times \overrightarrow {OB}$$ for this,

($$\because$$ $$\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a$$ and $$\overrightarrow a \times \overrightarrow a$$ = 0)

$$\overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 1 \cr 1 & { - 2} & 2 \cr } } \right|$$

$$= 6\widehat i - 3\widehat j - 6\widehat k = 3(2\widehat i - \widehat j - 2\widehat k)$$

So, $$\overrightarrow {OB} \times \overrightarrow {OC} = {{3\lambda } \over 2}(2\widehat i - \widehat j - 2\widehat k)$$

From Eq. (i), $$\left| {{{3\lambda } \over 2}\sqrt {4 + 1 + 4} } \right| = 9/2$$

$$\Rightarrow \left| {{{9\lambda } \over 2}} \right| = 9/2 \Rightarrow {9 \over 2}\left| \lambda \right| = {9 \over 2}$$

$$\Rightarrow \left| \lambda \right| = 1 \Rightarrow \lambda = \pm 1$$

But $$\lambda$$ > 0

$$\therefore$$ $$\lambda$$ = 1

So, $$\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \overrightarrow {OA} )$$ (By putting $$\lambda$$ = 1)

$$\Rightarrow \overrightarrow {OC} = {1 \over 2}( - \widehat i - 4\widehat j + \widehat k) = - {1 \over 2}\widehat i - 2\widehat j + {1 \over 2}\widehat k$$

Option (a)

Projection of $${\overrightarrow {OC} }$$ and $${\overrightarrow {OA} }$$ $$= {{\overrightarrow {OC} \,.\,\overrightarrow {OA} } \over {\left| {\overrightarrow {OA} } \right|}}$$

$$= {{{1 \over 2}( - 2 - 8 + 1)} \over 3} = {{ - 3} \over 2}$$

Option (b)

Area of $$\Delta OAB = {1 \over 2}\left| {\overrightarrow {OA} \times \overrightarrow {OB} } \right| = {1 \over 2}\left| {\overrightarrow {OA} } \right|\left. {\overrightarrow {OB} } \right|\sin 90^\circ$$

$$= {1 \over 2}\left| {3 \times 3} \right| = {9 \over 2}$$

Option (c)

Area of the $$\Delta ABC = {1 \over 2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|$$

$$= {1 \over 2}\left\| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 4} & 1 \cr { - 5/2} & { - 4} & {{{ - 1} \over 2}} \cr } } \right\|$$

$$= {1 \over 2}\left| {6\widehat i - 3\widehat j - 6\widehat k} \right| = {1 \over 2} \times 3 \times 3 = {9 \over 2}$$

Option (d)

The acute angle between the diagonals of the parallelogram with adjacent sides $${\overrightarrow {OA} }$$ and $${\overrightarrow {OC} }$$ = $$\theta$$

$$\Rightarrow {{(\overrightarrow {OA} + \overrightarrow {OC} )\,.\,(\overrightarrow {OA} - \overrightarrow {OC} )} \over {\left| {\overrightarrow {OA} + \overrightarrow {OC} } \right|\left| {\overrightarrow {OA} - \overrightarrow {OC} } \right|}} = \cos \theta$$

$$\Rightarrow \cos \theta = {{\left( {{3 \over 2}\widehat i + {3 \over 2}\widehat k} \right)\,.\,\left( {{5 \over 2}\widehat i + 4\widehat j + {1 \over 2}\widehat k} \right)} \over {{3 \over 2}\sqrt 2 \times \sqrt {{{90} \over 4}} }}$$

$$= {{18} \over {3\sqrt 2 \times \sqrt {90} }} = {1 \over {\sqrt 5 }} \ne {1 \over 2}$$ [$$\therefore$$ $$\theta$$ $$\ne$$ $$\pi$$/3]

2

### JEE Advanced 2020 Paper 2 Offline

MCQ (More than One Correct Answer)
Let a and b be positive real numbers. Suppose $$PQ = a\widehat i + b\widehat j$$ and $$PS = a\widehat i - b\widehat j$$ are adjacent sides of a parallelogram PQRS. Let u and v be the projection vectors of $$w = \widehat i + \widehat j$$ along PQ and PS, respectively. If |u| + |v| = |w| and if the area of the parallelogram PQRS is 8, then which of the following statements is/are TRUE?
A
a + b = 4
B
a $$-$$ b = 2
C
The length of the diagonal PR of the parallelogram PQRS is 4
D
w is an angle bisector of the vectors PQ and PS

## Explanation

Given vectors $$PQ = a\widehat i + b\widehat j$$ and $$PS = a\widehat i - b\widehat j$$ are adjacent sides of a parallelogram PQRS, so area of parallelogram PQRS =

|PQ $$\times$$ PS| = 2ab = 8 (given)

$$\therefore$$ ab = 4 ......(i)

According to the question,

|u| = |projection vector of w = $$\widehat i$$ + $$\widehat j$$ along PQ|

$$= \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right| = {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }}$$

and, similarly, |v| = |projection vector of $$w = \widehat i + \widehat j$$ along PS|

$$= \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right|$$

$$= {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} \Rightarrow |u| + |v|\, = \,|w|$$

$$\Rightarrow {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }} + {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} = \sqrt 2$$

$$\Rightarrow |a + b| + |a - b| = \sqrt 2 \sqrt {{a^2} + {b^2}}$$

If a $$\ge$$ b $$\ge$$ 0, then $$2a = \sqrt 2 \sqrt {{a^2} + {b^2}}$$

$$\Rightarrow 4{a^2} = 2{a^2} + 2{b^2} \Rightarrow {a^2} = {b^2}$$

$$\Rightarrow a = b$$ ....(ii)

From Eqs. (i) and (ii), we get

$$a = 2 = b \Rightarrow a + b = 4$$

and the length of diagonal PR

$$= |a\widehat i + b\widehat j + a\widehat i - b\widehat j|$$

$$= |2a\widehat i| = 2a = 4$$

And, the angle bisector of vector PQ and PS is along the vector

$$\pm {\lambda \over {\sqrt {{a^2} + {b^2}} }}((a\widehat i + b\widehat j) \pm (a\widehat i - b\widehat j))$$
3

### JEE Advanced 2020 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$\alpha$$2 + $$\beta$$2 + $$\gamma$$2 $$\ne$$ 0 and $$\alpha$$ + $$\gamma$$ = 1. Suppose the point (3, 2, $$-$$1) is the mirror image of the point (1, 0, $$-$$1) with respect to the plane $$\alpha$$x + $$\beta$$y + $$\gamma$$z = $$\delta$$. Then which of the following statements is/are TRUE?
A
$$\alpha$$ + $$\beta$$ = 2
B
$$\delta$$ $$-$$ $$\gamma$$ = 3
C
$$\delta$$ + $$\beta$$ = 4
D
$$\alpha$$ + $$\beta$$ + $$\gamma$$ = $$\delta$$

## Explanation

Since, the point A(3, 2, $$-$$1) is the mirror image of the point B(1, 0, $$-$$1) with respect to the plane $$\alpha$$x + $$\beta$$y + $$\gamma$$z = $$\delta$$, then

$${\alpha \over {3 - 1}} = {\beta \over {2 - 0}} = {\gamma \over {( - 1) - ( - 1)}}$$

$$\Rightarrow {\alpha \over 1} = {\beta \over 1} = {\gamma \over 0} \Rightarrow \gamma = 0$$

it is given that $$\alpha + \gamma = 1 \Rightarrow \alpha = 1$$, so $$\beta = 1$$.

And, the mid-point of AB, M(2, 1, $$-$$1) lies on the given plane, so

$$2\alpha + \beta - \gamma = \delta$$

$$\Rightarrow 2(1) + (1) - 0 = \delta \Rightarrow \delta = 3$$

$$\therefore$$ $$\alpha + \beta = 2$$, $$\delta - \gamma = 0$$, $$\delta + \beta = 4$$.
4

### JEE Advanced 2020 Paper 1 Offline

MCQ (More than One Correct Answer)
Let L1 and L2 be the following straight lines.

$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$ and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$.

Suppose the straight line

$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$

lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
A
$$\alpha$$ $$-$$ $$\gamma$$ = 3
B
l + m = 2
C
$$\alpha$$ $$-$$ $$\gamma$$ = 1
D
l + m = 0

## Explanation

Equation of given straight lines

$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$

and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$

having vector form respectively, are

and \eqalign{ & r = (\widehat i + \widehat k) + \lambda (\widehat i - \widehat j + 3\widehat k) \cr & r = (\widehat i + \widehat k) + v( - 3\widehat i - \widehat j + \widehat k) \cr}

$$\because$$ $$(\widehat i - \widehat j + 3\widehat k)\,.\,( - 3\widehat i - \widehat j + \widehat k)$$

= $$-$$3 + 1 + 3 = 1 is positive,

$$\therefore$$ Angle between supporting line vectors of lines L1 and L2 is acute, and point of intersection of given lines L1 and L2 is (1, 0, 1)

Now, vector along the acute angle bisector of vectors $$(\widehat i - \widehat j + 3\widehat k)$$ and $$( - 3\widehat i - \widehat j + \widehat k)$$ is $$(\widehat i - \widehat j + 2\widehat k)$$ or $$(\widehat i + \widehat j - 2\widehat k)$$

It is given that line

$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$ is the bisector of the acute angle between the lines L1 and L2, so

l = 1 and m = 1

and $${{I - \alpha } \over 1} = {{0 - 1} \over 1} = {{1 - \gamma } \over { - 2}}$$

$$\Rightarrow \alpha = 2,\,\gamma = - 1$$

$$\therefore$$ $$\alpha - \gamma = 3,\,l + m = 2$$

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