"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 :"

Question

Accepted Answer

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$P = (-2, -1, 3)$

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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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 :

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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 :

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