Let P be the image of the point Q(7, –2, 5) in the line and R(5, p, q) be a point on L. Then the square of the area of is __________. [2025]

Question

Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

Smart Achievers · Accepted Answer

(957)

Given, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be on the line.

Here P be the image of point.

Since, R is on the line L, then

$R \equiv (2 λ + 1, 3 λ - 1, 4 λ) = (5, p, q) [Given]$

$\Rightarrow 2 λ + 1 = 5$

$\Rightarrow 2 λ = 4 \Rightarrow λ = 2$

$∴ R \equiv (5, 5, 8)$

Since, T is also on the line L, then

$T \equiv (2 μ + 1, 3 μ - 1, 4 μ)$

Now, $\vec{Q T} = (2 μ - 6) \hat{i} + (3 μ + 1) \hat{j} + (4 μ - 5) \hat{k}$

and $\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ (Normal)

Taking $\vec{Q T} \cdot \vec{b} = 0$

$\Rightarrow 4 μ - 12 + 9 μ + 3 + 16 μ - 20 = 0 \Rightarrow 29 μ - 29 = 0$

$\Rightarrow μ = 1$

$∴ T \equiv (3, 2, 4)$

$\Rightarrow Q T = \sqrt{{(7 - 3)}^{2} + {(- 2 - 2)}^{2} + {(5 - 4)}^{2}}$

$= \sqrt{16 + 16 + 1} = \sqrt{33} \Rightarrow P Q = 2 \sqrt{33}$

Similarly, $R T = \sqrt{29}$

$\Rightarrow Area of △ P Q R = \frac{1}{2} \times \sqrt{29} \times 2 \sqrt{33} = \sqrt{29} \times \sqrt{33}$

$\Rightarrow {(Area of △ P Q R)}^{2} = {(\sqrt{29} \times \sqrt{33})}^{2} = 29 \times 33 = 957$ .

Solution

Q.
Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

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Solution

Q. Let P be the image of the point Q(7, –2, 5) in the line L:x–12=y+13=z4 and R(5, p, q) be a point on L. Then the square of the area of ∆PQR is __________. [2025]

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Q.
Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]