Let the position vectors of two points $P$ and $Q$ be $3 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+2 \hat{j}-4 \hat{k}$ respectively. Let $R$ and S be two points such that the direction rations of lines PR and QS are $(4,-1,2)$ and $(-2,1,-2)$, respectively. Let lines PR and QS intersect at T. If the vector $\overrightarrow{\mathrm{TA}}$ is perpendicular to both $\overrightarrow{\mathrm{PR}}$ and $\overrightarrow{\mathrm{QS}}$ and the length of vector $\overrightarrow{\mathrm{TA}}$ is $\sqrt{5}$ units, then the modulus of a position vector of A is :

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Three Dimensional Geometry

Let the position vectors of two points

P

and

Q

3 \hat{i} - \hat{j} + 2 \hat{k}

and

\hat{i} + 2 \hat{j} - 4 \hat{k}

respectively. Let

R

and S be two points such that the direction rations of lines PR and QS are

(4, - 1, 2)

and

(- 2, 1, - 2)

, respectively. Let lines PR and QS intersect at T. If the vector

TA

is perpendicular to both

PR

and

QS

and the length of vector

TA

5

units, then the modulus of a position vector of A is :

Answer: (D)

A point of

P R

(3 + 4 λ, - 1 - λ, 2 + 2 λ)

and a point of

QS

(1 - 2 μ, 2 + μ, - 4 - 2 μ)

. The second coordinates give

μ = - 3 - λ

, and then the first give

3 + 4 λ = 7 + 2 λ

, so

λ = 2

and

μ = - 5

; the third coordinates agree, both equal to

6

. Hence

T = (11, - 3, 6)

A vector perpendicular to both directions is

(4, - 1, 2) \times (- 2, 1, - 2) = (0, 4, 2)

with unit vector

\frac{1}{5} (0, 2, 1)

, so

T A = \pm 5 \cdot \frac{1}{5} (0, 2, 1) = \pm (0, 2, 1)

and

A

(11, - 1, 7)

(11, - 5, 5)

. In either case

O A = 121 + 1 + 49 = 121 + 25 + 25 = 171

The modulus of the position vector of

A

171

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option