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 $R(1,3,3)$ is 5, then the square of the area of $\Delta PQR$ is equal to:
- (1) 136
- (2) 140
- (3) 144
- (4) 148
Solution:
The equation of the line passing through $P(-2,-1,3)$ and parallel to $\vec{v} = 3\hat{i}+2\hat{j}+2\hat{k}$ is given by:
$$ \vec{r} = (-2\hat{i}-\hat{j}+3\hat{k}) + \lambda(3\hat{i}+2\hat{j}+2\hat{k}) $$
Any general point $Q$ on this line is:
$$ Q(3\lambda-2, 2\lambda-1, 2\lambda+3) $$
Given the distance $QR = 5$, where $R(1,3,3)$:
$$ QR^2 = 25 $$
$$ (3\lambda-2-1)^2 + (2\lambda-1-3)^2 + (2\lambda+3-3)^2 = 25 $$
$$ (3\lambda-3)^2 + (2\lambda-4)^2 + (2\lambda)^2 = 25 $$
Expanding the squares:
$$ 9(\lambda^2-2\lambda+1) + 4(\lambda^2-4\lambda+4) + 4\lambda^2 = 25 $$
$$ 9\lambda^2 – 18\lambda + 9 + 4\lambda^2 – 16\lambda + 16 + 4\lambda^2 = 25 $$
$$ 17\lambda^2 – 34\lambda + 25 = 25 $$
$$ 17\lambda^2 – 34\lambda = 0 \Rightarrow 17\lambda(\lambda-2) = 0 $$
Since $P$ and $Q$ are distinct points, $\lambda \neq 0$. Therefore, $\lambda = 2$.
Substituting $\lambda = 2$ to find coordinates of $Q$:
$$ Q = (3(2)-2, 2(2)-1, 2(2)+3) = (4, 3, 7) $$
Now, we find vectors $\vec{PQ}$ and $\vec{PR}$:
$$ \vec{PQ} = 6\hat{i} + 4\hat{j} + 4\hat{k} $$
$$ \vec{PR} = (1-(-2))\hat{i} + (3-(-1))\hat{j} + (3-3)\hat{k} = 3\hat{i} + 4\hat{j} $$
Area of $\Delta PQR = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$
$$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \end{vmatrix} $$
$$ = \hat{i}(0-16) – \hat{j}(0-12) + \hat{k}(24-12) $$
$$ = -16\hat{i} + 12\hat{j} + 12\hat{k} $$
Magnitude squared:
$$ |\vec{PQ} \times \vec{PR}|^2 = (-16)^2 + (12)^2 + (12)^2 = 256 + 144 + 144 = 544 $$
Square of the Area:
$$ \text{Area}^2 = \left( \frac{1}{2} \sqrt{544} \right)^2 = \frac{1}{4} (544) = 136 $$
Ans. (1)