Let the lines \(\mathrm{L}_{1}: \overrightarrow{\mathrm{r}}=\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}), \lambda \in \mathrm{R}\) \(L_{2}: \vec{r}=(\hat{i}+3 \hat{j}+\hat{k})+\mu(\hat{i}+\hat{j}+5 \hat{k}) ; \mu \in R\) intersect at the point S. If a plane \(\mathrm{ax}+\) by \(-z\) \(+\mathrm{d}=0\) passes through S and is parallel to both the lines \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\), then the value of \(\mathrm{a}+\mathrm{b}+\) d is equal to ____

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Three Dimensional Geometry

Let the lines

L_{1} : r = λ (\hat{i} + 2 \hat{j} + 3 \hat{k}), λ \in R

L_{2} : r = (\hat{i} + 3 \hat{j} + \hat{k}) + μ (\hat{i} + \hat{j} + 5 \hat{k}); μ \in R

intersect at the point S. If a plane

ax +

- z

+ d = 0

passes through S and is parallel to both the lines

L_{1}

and

L_{2}

, then the value of

a + b +

d is equal to ____

Answer: 5

Since the plane is parallel to both lines

L_{1}

and

L_{2}

, its normal vector must be perpendicular to both direction vectors

d_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}

and

d_{2} = \hat{i} + \hat{j} + 5 \hat{k}

.

The normal vector is:

n = d_{1} \times d_{2} = \hat{i} 11 \hat{j} 21 \hat{k} 35

= \hat{i} (10 - 3) - \hat{j} (5 - 3) + \hat{k} (1 - 2)

= 7 \hat{i} - 2 \hat{j} - \hat{k}

Since both lines pass through the origin (or lie in a plane through the origin), and the plane equation is

a x + b y - z + d = 0

, comparing with the normal vector

(7, - 2, - 1)

, we get

a = 7

b = - 2

, and the coefficient of

z

- 1

.

Since both lines pass through the origin,

d = 0

.

Therefore:

a + b + d = 7 + (- 2) + 0 = 5

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer