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1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
Let $$\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$$, $$\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$$ and $$\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$$ be three vectors such that, $$\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3$$ and $${\overrightarrow a }$$ is perpendicular to $${\overrightarrow b }$$. Then the greatest amongst the values of $${\left| {\overrightarrow a } \right|^2}$$ is _____________.

## Explanation

Since, $$\overrightarrow a .\,\overrightarrow b = 0$$

$$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$$ .... (1)

Also,

$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$

$$\Rightarrow$$ 5$$\beta$$2 + 30$$\beta$$ + 40 = 0

$$\Rightarrow$$ $$\beta$$ = $$-$$4, $$-$$2

$$\Rightarrow$$ $$\alpha$$ = 4, 8

$$\Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$$
2

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
Let Q be the foot of the perpendicular from the point P(7, $$-$$2, 13) on the plane containing the lines $${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$$ and $${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$$. Then (PQ)2, is equal to ___________.

## Explanation

Containing the line $$\left| {\matrix{ {x + 1} & {y - 1} & {z - 3} \cr 6 & 7 & 8 \cr 3 & 5 & 7 \cr } } \right| = 0$$

$$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$$

$$x - 2y + z = 0$$

$$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6$$

$$P{Q^2} = 96$$
3

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
If the projection of the vector $$\widehat i + 2\widehat j + \widehat k$$ on the sum of the two vectors $$2\widehat i + 4\widehat j - 5\widehat k$$ and $$- \lambda \widehat i + 2\widehat j + 3\widehat k$$ is 1, then $$\lambda$$ is equal to __________.

## Explanation

$$\overrightarrow a = \widehat i + 2\widehat j + \widehat k$$

$$\overrightarrow b = (2 - \lambda )\widehat i + 6\widehat j - 2\widehat k$$

$${{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = 1,\overrightarrow a \,.\,\overrightarrow b = 12 - \lambda$$

$$\left( {\overrightarrow a \,.\,\overrightarrow b } \right) = |\overrightarrow b {|^2}$$

$$\lambda$$2 $$-$$ 24$$\lambda$$ + 144 = $$\lambda$$2 $$-$$ 4$$\lambda$$ + 4 + 40

20$$\lambda$$ = 100 $$\Rightarrow$$ $$\lambda$$ = 5
4

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
Let the line L be the projection of the line $${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$$ in the plane x $$-$$ 2y $$-$$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d2 is equal to _______________.

## Explanation

L1 : $${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$$

for foot of $$\bot$$ r of (1, 3, 4) on x $$-$$ 2y $$-$$ z $$-$$ 3 = 0

(1 + t) $$-$$ 2(3 $$-$$ 2t) $$-$$ (4 $$-$$ t) $$-$$ 3 = 0

$$\Rightarrow$$ t = 2

So foot of $$\bot$$ r $$\buildrel \Delta \over =$$ (3, $$-$$1, 2) & point of intersection of L1 with plane is ($$-$$11, $$-$$3, $$-$$8)

dr's of L is <14, 2, 10>

$$\cong$$ <7, 1, 5>

Image

$$d = AB\sin \theta = |{{\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & { - 4} \cr 7 & 1 & 5 \cr } } \right|} \over {\sqrt {{7^2} + {1^2} + {5^2}} }}|$$

$$\Rightarrow {d^2} = {{{1^2} + {{(43)}^2} + {{(10)}^2}} \over {49 + 1 + 25}} = 26$$

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