Let P be the foot of the perpendicular from the point $Q(10, -3, -1)$ on the line $\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2}$. Then the area of the right angled triangle PQR, where R is the point $(3, -2, 1)$, is
- (1) $9\sqrt{15}$
- (2) $\sqrt{30}$
- (3) $8\sqrt{15}$
- (4) $3\sqrt{30}$
Solution:
Any point on the line is $P(7\lambda+3, -\lambda+2, -2\lambda-1)$.
Direction ratios of the line are $\vec{d} = <7, -1, -2>$.
Vector $\vec{QP} = P – Q = <7\lambda-7, -\lambda+5, -2\lambda>$.
Since $\vec{QP} \perp \text{Line}$, their dot product is 0:
$$7(7\lambda-7) – 1(-\lambda+5) – 2(-2\lambda) = 0$$
$$49\lambda – 49 + \lambda – 5 + 4\lambda = 0$$
$$54\lambda – 54 = 0 \implies \lambda = 1$$
Coordinates of $P$: $(10, 1, -3)$.
Now, find vectors for area calculation:
$$\vec{PQ} = Q – P = <0, -4, 2>$$
$$\vec{PR} = R – P = <3-10, -2-1, 1-(-3)> = <-7, -3, 4>$$
Area of $\Delta PQR = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$:
$$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 2 \\ -7 & -3 & 4 \end{vmatrix}$$
$$= \hat{i}(-16+6) – \hat{j}(0+14) + \hat{k}(0-28) = -10\hat{i} – 14\hat{j} – 28\hat{k}$$
Magnitude $|\vec{PQ} \times \vec{PR}| = \sqrt{100 + 196 + 784} = \sqrt{1080} = \sqrt{36 \times 30} = 6\sqrt{30}$.
$$Area = \frac{1}{2} \times 6\sqrt{30} = 3\sqrt{30}$$
Ans.(4)