Question

Let $L_1$ be the line of intersection of the planes given by the equations $2x + 3y + z = 4$ and $x + 2y + z = 5$.

Let $L_2$ be the line passing through the point $P(2, -1, 3)$ and parallel to $L_1$. Let $M$ denote the plane given by the equation

$2x + y - 2z = 6$.

Suppose that the line $L_2$ meets the plane $M$ at the point $Q$. Let $R$ be the foot of the perpendicular drawn from $P$ to the plane $M$.

Then which of the following statements is (are) TRUE?

• A. The length of the line segment $PQ$ is $9\sqrt{3}$
• B. The length of the line segment $QR$ is $15$
• C. The area of $\triangle PQR$ is $\frac{3}{2}\sqrt{234}$
• D. The acute angle between the line segments $PQ$ and $PR$ is $cos^{-1}(\frac{1}{2\sqrt{3}})$

Answer 1

Answer: (A), (C)

(A), (C)

Answer 2

The line $L_1$ is the intersection of the planes $2x + 3y + z = 4$ and $x + 2y + z = 5$. The direction vector of $L_1$ is the cross product of the normal vectors of the two planes: $\vec{v_1} = (2, 3, 1) \times (1, 2, 1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = (3-2)\mathbf{i} - (2-1)\mathbf{j} + (4-3)\mathbf{k} = (1, -1, 1)$.

The line $L_2$ passes through $P(2, -1, 3)$ and is parallel to $L_1$. So, the direction vector of $L_2$ is $(1, -1, 1)$. The parametric equation of $L_2$ is $\mathbf{r}(s) = (2, -1, 3) + s(1, -1, 1) = (2+s, -1-s, 3+s)$.

The plane $M$ is given by $2x + y - 2z = 6$. The point $Q$ is the intersection of $L_2$ and plane $M$. Substitute the coordinates of a point on $L_2$ into the equation of $M$: $2(2+s) + (-1-s) - 2(3+s) = 6$ $4 + 2s - 1 - s - 6 - 2s = 6$ $-3 - s = 6 \implies s = -9$. Substitute $s = -9$ into the equation of $L_2$ to find $Q$: $Q = (2+(-9), -1-(-9), 3+(-9)) = (-7, 8, -6)$.

The length of $PQ$ is the distance between $P(2, -1, 3)$ and $Q(-7, 8, -6)$: $PQ = \sqrt{(-7-2)^2 + (8-(-1))^2 + (-6-3)^2} = \sqrt{(-9)^2 + 9^2 + (-9)^2} = \sqrt{81 + 81 + 81} = \sqrt{3 \cdot 81} = 9\sqrt{3}$. Statement (A) is TRUE.

$R$ is the foot of the perpendicular from $P(2, -1, 3)$ to the plane $M: 2x + y - 2z = 6$. The line through $P$ perpendicular to $M$ has the direction vector equal to the normal vector of $M$, $\vec{n_M} = (2, 1, -2)$. The parametric equation of this line is $\mathbf{r}(t) = (2, -1, 3) + t(2, 1, -2) = (2+2t, -1+t, 3-2t)$. Substitute the coordinates into the equation of $M$ to find $R$: $2(2+2t) + (-1+t) - 2(3-2t) = 6$ $4 + 4t - 1 + t - 6 + 4t = 6$ $-3 + 9t = 6 \implies 9t = 9 \implies t = 1$. Substitute $t = 1$ to find $R$: $R = (2+2(1), -1+1, 3-2(1)) = (4, 0, 1)$.

The length of $QR$ is the distance between $Q(-7, 8, -6)$ and $R(4, 0, 1)$: $QR = \sqrt{(4-(-7))^2 + (0-8)^2 + (1-(-6))^2} = \sqrt{11^2 + (-8)^2 + 7^2} = \sqrt{121 + 64 + 49} = \sqrt{234}$. Statement (B) says $QR = 15$, which is false.

The area of $\triangle PQR$ can be calculated using the coordinates of $P, Q, R$. Since $R$ is the foot of the perpendicular from $P$ to the plane containing $Q$ and $R$, the line segment $PR$ is perpendicular to the plane $M$. The line segment $QR$ lies in the plane $M$. Therefore, $PR$ is perpendicular to $QR$. $\triangle PQR$ is a right-angled triangle with the right angle at $R$. The area of $\triangle PQR = \frac{1}{2} \cdot QR \cdot PR$. We have $QR = \sqrt{234}$. The length of $PR$ is the distance between $P(2, -1, 3)$ and $R(4, 0, 1)$: $PR = \sqrt{(4-2)^2 + (0-(-1))^2 + (1-3)^2} = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. Area of $\triangle PQR = \frac{1}{2} \cdot \sqrt{234} \cdot 3 = \frac{3}{2}\sqrt{234}$. Statement (C) is TRUE.

The acute angle between $PQ$ and $PR$ is the angle $\angle RPQ$. In the right-angled triangle $\triangle PQR$, this angle is $\angle P$. We have $PR = 3$ (adjacent side) and $PQ = 9\sqrt{3}$ (hypotenuse). $\cos(\angle RPQ) = \frac{PR}{PQ} = \frac{3}{9\sqrt{3}} = \frac{1}{3\sqrt{3}}$. Statement (D) says the acute angle is $\cos^{-1}(\frac{1}{2\sqrt{3}})$, which is false.

The true statements are (A) and (C).