Question

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:

• A. \( 9\sqrt{15} \)
• B. \( \sqrt{30} \)
• C. \( 8\sqrt{15} \)
• D. \( 3\sqrt{30} \)

Hint:

When finding the area of a triangle using vectors, compute the determinant of a 3×3 matrix formed by the two vectors.

Answer 1

The Correct Option is D

To find the area of the right-angled triangle PQR, where \( R(3,-2,1) \) and \( Q(10,-3,-1) \), we first need to find point \( P \), the foot of the perpendicular from \( Q \) to the line:

\[ \frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2} \]

This line can be expressed in vector form as: 
\(\mathbf{r} = \langle3,2,-1\rangle + \lambda\langle7,-1,-2\rangle\).

The direction vector of the line is \(\mathbf{a} = \langle7,-1,-2\rangle\), point \( Q \) is \(\langle10,-3,-1\rangle\), and a point on the line can be taken as \(\langle3,2,-1\rangle\).

The vector \(\mathbf{PQ}\) is perpendicular to the line's direction. For PQ calculated at point \(\langle x,y,z\rangle\) (on the line),

\(\mathbf{PQ}=\langle x-10,y+3,z+1\rangle\).

Since \(\mathbf{PQ}\) is perpendicular to the line's direction vector:

\[\langle x-10,y+3,z+1\rangle\cdot\langle7,-1,-2\rangle=0\]

Substitute \(\mathbf{r} = \langle3+7\lambda,2-\lambda,-1-2\lambda\rangle\) from the line equation:

\[\langle3+7\lambda-10,2-\lambda+3,-1-2\lambda+1\rangle\cdot\langle7,-1,-2\rangle=0\]

Simplifying,

\[\langle7\lambda-7,-\lambda+5,-2\lambda\rangle\cdot\langle7,-1,-2\rangle=0\]

Expanding, \(49\lambda-49+\lambda-5+4\lambda=0\). Further simplify:

\[54\lambda-54=0\Rightarrow \lambda=1\]

Substitute back to find \(P\):

\(x=3+7\times1=10,\ y=2-1=1,\ z=-1-2=-3\), so \(P(10,1,-3)\).

To find the area of triangle PQR:

Vectors: \(\mathbf{PQ}=\langle0,-4,-2\rangle\) and \(\mathbf{PR}=\langle-7,-3,4\rangle\).

The cross product, \(\mathbf{PQ}\times\mathbf{PR}\):

\(\det\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&-4&-2\\-7&-3&4\end{vmatrix}\)=\[ \mathbf{i}(-4\times4+2\times3)-\mathbf{j}(0\times4+2\times7)+\mathbf{k}(0\times(-3)+4\times7)\] = \(\mathbf{i}(-10)-\mathbf{j}(14)+\mathbf{k}(28)\)

=\(\langle-10,-14,28\rangle\)

The magnitude is:

\(|\langle-10,-14,28\rangle|=\sqrt{(-10)^2+(-14)^2+28^2}\) = \(\sqrt{100+196+784} = \sqrt{1080}\)

Area of triangle: \( \frac{1}{2}|\mathbf{PQ}\times\mathbf{PR}|=\frac{\sqrt{1080}}{2}=3\sqrt{30}\)

Thus, the area is \(3\sqrt{30} \).