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1

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

Numerical
Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $$-$$ y + z + 3 = 0 and let R(3, 5, $$\gamma$$) be a point of this plane. Then the square of the length of the line segment SR is ___________.

## Explanation Since R(3, 5, $$\gamma$$) lies on the plane 2x $$-$$ y + z + 3 = 0.

Therefore, 6 $$-$$ 5 + $$\gamma$$ + 3 = 0

$$\Rightarrow$$ $$\gamma$$ = $$-$$4

Now,

dr's of line QS are 2, $$-$$1, 1

equation of line QS is

$${{x - 1} \over 2} = {{y - 3} \over { - 1}} = {{z - 4} \over 1} = \lambda$$ (say)

$$\Rightarrow F(2\lambda + 1, - \lambda + 3,\lambda + 4)$$

F lies in the plane

$$\Rightarrow 2(2\lambda + 1) - ( - \lambda + 3) + (\lambda + 4) + 3$$ = 0

$$\Rightarrow 4\lambda + 2 + \lambda - 3 + \lambda + 7 = 0$$

$$\Rightarrow 6\lambda + 6 = 0 \Rightarrow \lambda = - 1$$

$$\Rightarrow$$ F($$-$$1, 4, 3)

Since, F is mid-point of QS.

Therefore, coordinated of S are ($$-$$3, 5, 2).

So, SR = $$\sqrt {36 + 0 + 36} = \sqrt {72}$$

SR2 = 72.
2

### 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

$$\Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$$
3

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

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

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

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

$$P{Q^2} = 96$$
4

### 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

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