Let d be the distance of the point of intersection of the lines and from the point(7, 8, 9). Then is equal to [2024]

Question

Let d be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point(7, 8, 9). Then $d^{2} + 6$ is equal to [2024]

Smart Achievers · Accepted Answer

(2)

Let $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = λ$

$\Rightarrow x = 3 λ - 6, y = 2 λ, z = λ - 1$

and let $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = μ$

$\Rightarrow x = 4 μ + 7, y = 3 μ + 9, z = 2 μ + 4$

Since the given lines intersect so we have

$3 λ - 6 = 4 μ + 7, 2 λ = 3 μ + 9, λ - 1 = 2 μ + 4$

$\Rightarrow 3 λ - 4 μ = 13 a n d λ = 2 μ + 5$

$\Rightarrow 3 (2 μ + 5) - 4 μ = 13 \Rightarrow 2 μ = - 2$

$\Rightarrow μ = - 1 a n d λ = 3$

So, point of intersection is (3, 6, 2)

$∴ d = \sqrt{{(3 - 7)}^{2} + {(6 - 8)}^{2} + {(2 - 9)}^{2}}$

$= \sqrt{16 + 4 + 49} = \sqrt{69}$

$∴ d^{2} + 6 = 69 + 6 = 75$ .

Solution

Q.
Let d be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point(7, 8, 9). Then $d^{2} + 6$ is equal to [2024]

1 72

2 75

3 78

4 69

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