The distance of the point Q ( 0 , 2 , – 2 ) form the line passing through the point P ( 5 , – 4 , 3 ) and perpendicular to the lines r → = − 3 i ^ + 2 k ^ + λ 2 i ^ + 3 j ^ + 5 k ^ , λ ∈ ℝ and r → = i ^ − 2 j ^ + k ^ + μ − i ^ + 3 j ^ + 2 k ^ , μ ∈ ℝ is

Question

Aicharya · Accepted Answer

Plotting the diagram of the given data we get,

Now, let $a, b, c$ be the DR's of the line perpendicular to $(- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k}) and (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k})$ .

$\Rightarrow 2 a + 3 b + 5 c = 0, - a + 3 b + 2 c = 0$

Now, solving $2 a + 3 b + 5 c = 0 & - a + 3 b + 2 c = 0$ we get,

$\Rightarrow \frac{a}{- 9} = \frac{b}{- 9} = \frac{c}{9}$

$\Rightarrow \frac{a}{1} = \frac{b}{1} = \frac{c}{- 1}$

So, the equation will be,

$\Rightarrow \frac{x - 5}{1} = \frac{y + 4}{1} = \frac{z - 3}{- 1} . . . (i)$

Let $R$ be the foot of perpendicular from $Q (0, 2, - 2) to (i)$ .

$\Rightarrow \frac{x - 5}{1} = \frac{y + 4}{1} = \frac{z - 3}{- 1} = k$

$\Rightarrow x = k + 5, y = k - 4, z = - k + 3$

$\Rightarrow R \equiv (k + 5, k - 4, - k + 3)$

So, DR's of $Q R$ are $(k + 5, k - 6, 5 - k)$ .

Now, using perpendicular condition we get,

$\Rightarrow 1 (k + 5) + 1 (k - 6) + (- 1) (5 - k) = 0$

$\Rightarrow k = 2$

$\Rightarrow R \equiv (7, - 2, 1)$

$\Rightarrow Q R = \sqrt{49 + 16 + 9}$

$\Rightarrow Q R = \sqrt{74}$

The distance of the point Q ( 0 , 2 , – 2 ) form the line passing through the point P ( 5 , – 4 , 3 ) and…

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