Question 79

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 :

Solution

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Line 1: $$\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1} = s$$. Point: $$(-6+3s, 2s, -1+s)$$.

Line 2: $$\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2} = t$$. Point: $$(7+4t, 9+3t, 4+2t)$$.

At intersection:

$$ -6+3s = 7+4t \quad \Rightarrow 3s - 4t = 13 \quad ... (1) $$ $$ 2s = 9+3t \quad \Rightarrow 2s - 3t = 9 \quad ... (2) $$ $$ -1+s = 4+2t \quad \Rightarrow s - 2t = 5 \quad ... (3) $$

From (3): $$s = 5 + 2t$$. Substitute in (2): $$2(5+2t) - 3t = 9 \Rightarrow 10 + t = 9 \Rightarrow t = -1$$.

$$s = 5 - 2 = 3$$.

Check (1): $$9 - (-4) = 13$$ âś“.

Intersection point: $$(-6+9, 6, -1+3) = (3, 6, 2)$$.

Distance from $$(3,6,2)$$ to $$(7,8,9)$$:

$$ d = \sqrt{16 + 4 + 49} = \sqrt{69} $$ $$ d^2 = 69 $$ $$ d^2 + 6 = 75 $$

The correct answer is Option (4): 75.

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