The distance of the point $(-1,2,3)$ from the plane $\overrightarrow{\mathrm{r}} .(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=10$ parallel to the line of the shortest distance between the lines $\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$ and $\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$ is :

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Three Dimensional Geometry

The distance of the point

(- 1, 2, 3)

from the plane

r . (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 10

parallel to the line of the shortest distance between the lines

r = (\hat{i} - \hat{j}) + λ (2 \hat{i} + \hat{k})

and

r = (2 \hat{i} - \hat{j}) + μ (\hat{i} - \hat{j} + \hat{k})

is :

Answer: (C)

The direction of the shortest distance between the two skew lines is

n = b_{1} \times b_{2} = (2 \hat{i} + \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = \hat{i} - \hat{j} - 2 \hat{k}

. The line through

(- 1, 2, 3)

parallel to

n

\frac{x + 1}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{- 2} = r

. Parametrically,

(x, y, z) = (r - 1, 2 - r, 3 - 2 r)

. Substitute into the plane equation

x - 2 y + 3 z = 10

(r - 1) - 2 (2 - r) + 3 (3 - 2 r) = 10

gives

r = - 2

. So the point on the plane is

Q (- 3, 4, 7)

. Distance

P Q = 4 + 4 + 16 = 26

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option