Q.24. Let the image of the point P(0, -5, 0) in the line be the point R and the image of the point

Q: What is the correct answer for this exam Mathematics Q24?

The correct answer is 162. Given point P(0, -5, 0).Line $L_1: \frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}$.Let M be the foot of the perpendicular from P to $L_1$. Any point on $L_1$ is $(2k+1, k, -2k-1)$.Vector PM is $(2k+1-0, k-(-5), -2k-1-0) = (2k+1, k+5, -2k-1)$.Direction vector of $L_1$ is $\vec{d_1} = (2,1,-2)$.Since PM is perpendicular to $L_1$, $\vec{PM} \cdot \vec{d_1} = 0$.$2(2k+1) + 1(k+5) - 2(-2k-1) = 0$.$4k+2 +

Q: Q.24. Let the image of the point P(0, -5, 0) in the line be the point R and the image of the point Q(0, -1/2, 0) in the line be the point S. Then the square of the area of the parallelogram PQRS is

The correct answer is 162. Given point P(0, -5, 0).Line $L_1: \frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}$.Let M be the foot of the perpendicular from P to $L_1$. Any point on $L_1$ is $(2k+1, k, -2k-1)$.Vector PM is $(2k+1-0, k-(-5), -2k-1-0) = (2k+1, k+5, -2k-1)$.Direction vector of $L_1$ is $\vec{d_1} = (2,1,-2)$.Since PM is perpendicular to $L_1$, $\vec{PM} \cdot \vec{d_1} = 0$.$2(2k+1) + 1(k+5) - 2(-2k-1) = 0$.$4k+2 +

Question

Q.24. Let the image of the point P(0, -5, 0) in the line  be the point R and the image of the point Q(0, -1/2, 0) in the line  be the point S. Then the square of the area of the parallelogram PQRS is

FirstInTest · Accepted Answer

Given point P(0, -5, 0).Line $L_1: \frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}$.Let M be the foot of the perpendicular from P to $L_1$. Any point on $L_1$ is $(2k+1, k, -2k-1)$.Vector PM is $(2k+1-0, k-(-5), -2k-1-0) = (2k+1, k+5, -2k-1)$.Direction vector of $L_1$ is $\vec{d_1} = (2,1,-2)$.Since PM is perpendicular to $L_1$, $\vec{PM} \cdot \vec{d_1} = 0$.$2(2k+1) + 1(k+5) - 2(-2k-1) = 0$.$4k+2 + k+5 + 4k+2 = 0$.$9k+9 = 0 \Rightarrow k=-1$.So M is $(2(-1)+1, -1, -2(-1)-1) = (-1, -1, 1)$.R is the image of P. M is the midpoint of PR.So $M = \frac{P+R}{2} \Rightarrow R = 2M - P$.$R = 2(-1,-1,1) - (0,-5,0) = (-2,-2,2) - (0,-5,0) = (-2,3,2)$.Given point Q(0, -1/2, 0).Line $L_2: \frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1}$.Let N be the foot of the perpendicular from Q to $L_2$. Any point on $L_2$ is $(-\lambda+1, 4\lambda-9, \lambda-1)$.Vector QN is $(-\lambda+1-0, 4\lambda-9-(-1/2), \lambda-1-0) = (-\lambda+1, 4\lambda-9+1/2, \lambda-1) = (-\lambda+1, 4\lambda-17/2, \lambda-1)$.Direction vector of $L_2$ is $\vec{d_2} = (-1,4,1)$.Since QN is perpendicular to $L_2$, $\vec{QN} \cdot \vec{d_2} = 0$.$-1(-\lambda+1) + 4(4\lambda-17/2) + 1(\lambda-1) = 0$.$\lambda-1 + 16\lambda-34 + \lambda-1 = 0$.$18\lambda - 36 = 0 \Rightarrow 18\lambda = 36 \Rightarrow \lambda=2$.So N is $(-2+1, 4(2)-9, 2-1) = (-1, -1, 1)$.S is the image of Q. N is the midpoint of QS.So $N = \frac{Q+S}{2} \Rightarrow S = 2N - Q$.$S = 2(-1,-1,1) - (0,-1/2,0) = (-2,-2,2) - (0,-1/2,0) = (-2, -3/2, 2)$.We have the vertices of the parallelogram PQRS:P(0, -5, 0)Q(0, -1/2, 0)R(-2, 3, 2)S(-2, -3/2, 2)Area of parallelogram PQRS is $|\vec{PQ} 	imes \vec{PS}|$.$\vec{PQ} = (0-0, -1/2 - (-5), 0-0) = (0, 9/2, 0)$.$\vec{PS} = (-2-0, -3/2 - (-5), 2-0) = (-2, -3/2 + 10/2, 2) = (-2, 7/2, 2)$.$\vec{PQ} 	imes \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 9/2 & 0 \ -2 & 7/2 & 2 \end{vmatrix}$$= \hat{i}((9/2)(2) - 0) - \hat{j}(0 - 0) + \hat{k}(0 - (9/2)(-2))$$= \hat{i}(9) - \hat{j}(0) + \hat{k}(9) = 9\hat{i} + 9\hat{k}$.Area $= |9\hat{i} + 9\hat{k}| = \sqrt{9^2+9^2} = \sqrt{81+81} = \sqrt{162}$.The square of the area is $(\sqrt{162})^2 = 162$.The solution in the image calculates PR and QS.Area of parallelogram = $\frac{1}{2} |\vec{PR} 	imes \vec{QS}|$.$\vec{PR} = (-2-0, 3-(-5), 2-0) = (-2, 8, 2)$.$\vec{QS} = (-2-0, -3/2 - (-1/2), 2-0) = (-2, -1, 2)$.$\vec{PR} 	imes \vec{QS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 8 & 2 \ -2 & -1 & 2 \end{vmatrix}$$= \hat{i}(16 - (-2)) - \hat{j}(-4 - (-4)) + \hat{k}(2 - (-16))$$= \hat{i}(18) - \hat{j}(0) + \hat{k}(18) = 18\hat{i} + 18\hat{k}$.$|\vec{PR} 	imes \vec{QS}| = \sqrt{18^2+18^2} = \sqrt{2 	imes 18^2} = 18\sqrt{2}$.Area $= \frac{1}{2} (18\sqrt{2}) = 9\sqrt{2}$.Square of area $= (9\sqrt{2})^2 = 81 	imes 2 = 162$.Both methods yield the same result. The solution in the image is correct.

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