Let a line l pass through the origin and be perpendicular to the lines, l 1 : r = ( i ^ - 11 j ^ - 7 k ^ ) + ( i ^ + 2 j ^ + 3 k ^ ) , and l 2 : r = ( - i ^ + k ^ ) + ? ( 2 i ^ + 2 j ^ + k ^ ) . If P is the point of intersection of l and l 1

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Solution

Q.
Let a line $l$ pass through the origin and be perpendicular to the lines, $l_{1} : \vec{r} = (\hat{i} - 11 \hat{j} - 7 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k}), λ \in ℝ$ and $l_{2} : \vec{r} = (- \hat{i} + \hat{k}) + μ (2 \hat{i} + 2 \hat{j} + \hat{k}), μ \in ℝ .$ If P is the point of intersection of $l$ and $l_{1}$ , and $Q (α, β, γ)$ is the foot of the perpendicular from $P$ on $l_{2}$ , then $9 (α + β + γ)$ is equal to _______ . [2023]

Ans.
(5)

$Let l = (0 \hat{i} + 0 \hat{j} + 0 \hat{k}) + γ (a \hat{i} + b \hat{j} + c \hat{k}) = γ (a \hat{i} + b \hat{j} + c \hat{k})$

Since $l$ is perpendicular to the lines $l_{1}$ and $l_{2}$ ,

$a \hat{i} + b \hat{j} + c \hat{k} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{matrix} |$ $= \hat{i} (2 - 6) - \hat{j} (1 - 6) + \hat{k} (2 - 4)$

$= - 4 \hat{i} + 5 \hat{j} - 2 \hat{k}$ $∴$ $l = γ (- 4 \hat{i} + 5 \hat{j} - 2 \hat{k})$

$P$ is the point of intersection of $l$ and $l_{1}$ .

$- 4 γ = 1 + λ, 5 γ = - 11 + 2 λ, - 2 γ = - 7 + 3 λ$

$By solving these equations, we get γ = - 1 .$

$So, P \equiv (4, - 5, 2)$

$Let Q (- 1 + 2 μ, 2 μ, 1 + μ)$

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

$\Rightarrow [(2 μ - 5) \hat{i} + (2 μ + 5) \hat{j} + (μ - 1) \hat{k}] \cdot (2 \hat{i} + 2 \hat{j} + \hat{k}) = 0$

$\Rightarrow 4 μ - 10 + 4 μ + 10 + μ - 1 = 0 \Rightarrow 9 μ = 1 \Rightarrow μ = \frac{1}{9}$

$Hence, 9 (α + β + γ) = 9 (\frac{- 7}{9} + \frac{2}{9} + \frac{10}{9}) = 5$

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