Let a line $l$ pass through the origin and be perpendicular to the lines $l_{1}: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-11 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}), \lambda \in \mathbb{R}$ and $l_{2}: \overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$. If P is the point of intersection of $l$ and $l_{1}$, and $\mathrm{Q}(\alpha, \beta, \gamma)$ is the foot of perpendicular from P on $l_{2}$, then $9(\alpha+\beta+\gamma)$ is equal to

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Three Dimensional Geometry

Let a line

l

pass through the origin and be perpendicular to the lines

l_{1} : r = (\hat{i} - 11 \hat{j} - 7 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k}), λ \in R

and

l_{2} : r = (- \hat{i} + \hat{k}) + μ (2 \hat{i} + 2 \hat{j} + \hat{k}), μ \in R

.

If P is the point of intersection of

l

and

l_{1}

, and

Q (α, β, γ)

is the foot of perpendicular from P on

l_{2}

, then

9 (α + β + γ)

is equal to

Answer: 5

The direction of

l

must be perpendicular to the directions of both

l_{1}

and

l_{2}

, so it is along

(\hat{i} + 2 \hat{j} + 3 \hat{k}) \times (2 \hat{i} + 2 \hat{j} + \hat{k}) = - 4 \hat{i} + 5 \hat{j} - 2 \hat{k}

. Since

l

passes through the origin, a general point on it is

t (- 4, 5, - 2)

. Equating this with a general point

(1 + λ, - 11 + 2 λ, - 7 + 3 λ)

l_{1}

gives

t = - 1

and

λ = 3

, so

P = (4, - 5, 2)

. A general point of

l_{2}

Q = (- 1 + 2 μ, 2 μ, 1 + μ)

, hence

P Q = (2 μ - 5, 2 μ + 5, μ - 1)

. For

Q

to be the foot of the perpendicular,

P Q \cdot (2 \hat{i} + 2 \hat{j} + \hat{k}) = 0

, which gives

9 μ - 1 = 0

, so

μ = \frac{1}{9}

. Then

Q = (- \frac{7}{9}, \frac{2}{9}, \frac{10}{9})

, and therefore

9 (α + β + γ) = 9 \cdot \frac{5}{9} = 5

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer