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Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3 \hat{i}+2 \hat{j}+2 \hat{k}$. If the distance of the point Q from the point $\mathrm{R}(1,3,3)$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :

JEE · Math · previous-year question

  1. A.148
  2. B.144
  3. C.136correct
  4. D.140

Answer

C. 136

Explanation

$$ P = (-2, -1, 3) $$ Since the line through $ P $ is parallel to the vector $$ \vec{d} = (3, 2, 2), $$ any point $ Q $ on this line can be expressed as: $$ Q = P + t\,\vec{d} = (-2 + 3t,\, -1 + 2t,\, 3 + 2t), $$ where $ t $ is a real number. Given that $ P $ and $ Q $ must be distinct, we require $ t \neq 0 $. The point $ R $ is given by: $$ R = (1, 3, 3). $$ The distance from $ Q $ to $ R $ is $ 5 $, so we have: $$ \left[ (-2 + 3t - 1)^2 + (-1 + 2t - 3)^2 + (3 + 2t - 3)^2 \right] = 5^2. $$ Simplify each coordinate difference: For the $ x $-coordinate: $$ -2 + 3t - 1 = 3t - 3 = 3(t - 1). $$ For the $ y $-coordinate: $$ -1 + 2t - 3 = 2t - 4 = 2(t - 2). $$ For the $ z $-coordinate: $$ 3 + 2t - 3 = 2t. $$ So the equation becomes: $$ [3(t - 1)]^2 + [2(t - 2)]^2 + (2t)^2 = 25. $$ Expanding the squares: $$ 9(t - 1)^2 + 4(t - 2)^2 + 4t^2 = 25. $$ Expand each term: $$ 9(t^2 - 2t + 1) + 4(t^2 - 4t + 4) + 4t^2 = 25, $$ which simplifies to: $$ 9t^2 - 18t + 9 + 4t^2 - 16t + 16 + 4t^2 = 25. $$ Combine like terms: $$ (9t^2 + 4t^2 + 4t^2) - (18t + 16t) + (9 + 16) = 25, $$ $$ 17t^2 - 34t + 25 = 25. $$ Subtract 25 from both sides: $$ 17t^2 - 34t = 0. $$ Factor out $ 17t $: $$ 17t(t - 2) = 0. $$ Since $ t \neq 0 $, we must have: $$ t = 2. $$ Substitute $ t = 2 $ back into the equation for $ Q $: $$ Q = (-2 + 3(2),\, -1 + 2(2),\, 3 + 2(2)) = (4, 3, 7). $$ Next, to find the square of the area of triangle $ PQR $, first compute the vectors: $$ \vec{PQ} = Q - P = (4 - (-2),\, 3 - (-1),\, 7 - 3) = (6, 4, 4), $$ $$ \vec{PR} = R - P = (1 - (-2),\, 3 - (-1),\, 3 - 3) = (3, 4, 0). $$ The area of the triangle is given by: $$ \text{Area} = \frac{1}{2} \, \| \vec{PQ} \times \vec{PR} \|. $$ Calculate the cross product: $$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \\ \end{vmatrix}. $$ Expanding the determinant: $ \hat{i} $-component: $$ 4 \times 0 - 4 \times 4 = -16, $$ $ \hat{j} $-component: $$ -(6 \times 0 - 4 \times 3) = 12, $$ $ \hat{k} $-component: $$ 6 \times 4 - 4 \times 3 = 24 - 12 = 12. $$ Thus, $$ \vec{PQ} \times \vec{PR} = (-16,\, 12,\, 12). $$ Find the magnitude squared of the cross product: $$ \| \vec{PQ} \times \vec{PR} \|^2 = (-16)^2 + 12^2 + 12^2 = 256 + 144 + 144 = 544. $$ The square of the area of triangle $ PQR $ is: $$ (\text{Area})^2 = \left(\frac{1}{2}\right)^2 \| \vec{PQ} \times \vec{PR} \|^2 = \frac{1}{4} \times 544 = 136. $$ Thus, the square of the area of $ \triangle PQR $ is $$136.$$

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