Let Q(a,b,c) be the image of the point P(3,2,1) in the line x - 1 1 = y 2 = z - 1 1 . Then the distance of Q from the line x - 9 3 = y - 9 2 = z - 5 - 2 is. [2026]

Question

Let Q(a,b,c) be the image of the point P(3,2,1) in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} .$ Then the distance of Q from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is. [2026]

Smart Achievers · Accepted Answer

(2)

$drs of PN = ⟨ r - 2, 2 r - 2, r ⟩$

$1 (r - 2) + 2 (2 r - 2) + 1 (r) = 0$

$6 r = 6 \Rightarrow r = 1$

$∴$ $N \equiv (2, 2, 2)$

$\Rightarrow Q \equiv (1, 2, 3)$

$A Q = \sqrt{64 + 49 + 4} = \sqrt{117}$

$A M = \frac{| 24 + 14 - 4 |}{\sqrt{9 + 4 + 4}} = \frac{34}{\sqrt{17}} = 2 \sqrt{17}$

$∴$ $Q M = \sqrt{117 - 68} = \sqrt{49} = 7$

Solution

Q.
Let Q(a,b,c) be the image of the point P(3,2,1) in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} .$ Then the distance of Q from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is. [2026]

1 5

2 7

3 8

4 6

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Solution

Q. Let Q(a,b,c) be the image of the point P(3,2,1) in the line x-11=y2=z-11. Then the distance of Q from the line x-93=y-92=z-5-2 is. [2026]

1 5

2 7

3 8

4 6

Ans. (2) drs of PN=⟨r-2, 2r-2, r⟩ 1(r-2)+2(2r-2)+1(r)=0 6r=6⇒r=1 ∴ N≡(2,2,2) ⇒Q≡(1,2,3) AQ=64+49+4=117 AM=|24+14-4|9+4+4=3417=217 ∴ QM=117-68=49=7

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Q.
Let Q(a,b,c) be the image of the point P(3,2,1) in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} .$ Then the distance of Q from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is. [2026]