Question ID: #874
Let a line L passing through the point $P(1,1,1)$ be perpendicular to the lines $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$ and $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point $S(1,0,-1)$ intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to
Solution:
Direction ratios of given lines are $\vec{d_1} = <1, 1, 0>$ and $\vec{d_2} = <4, 1, 1>$.
Since line L is perpendicular to both, its direction $\vec{d}$ is along $\vec{d_1} \times \vec{d_2}$.
$$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 4 & 1 & 1 \end{vmatrix} = \hat{i}(1-0) – \hat{j}(1-0) + \hat{k}(1-4)$$
$$\vec{d} = \hat{i} – \hat{j} – 3\hat{k}$$
Direction ratios of L are $<1, -1, -3>$.
Equation of line L passing through $P(1,1,1)$:
$$\frac{x-1}{1} = \frac{y-1}{-1} = \frac{z-1}{-3} = \lambda$$
Intersection with yz-plane (Put $x=0$):
$$\frac{-1}{1} = \lambda \Rightarrow \lambda = -1$$
$$y = 1 – (-1) = 2, \quad z = 1 – 3(-1) = 4$$
So, $Q(0, 2, 4)$.
Another line parallel to L passes through $S(1,0,-1)$. Direction is same $<1, -1, -3>$.
Equation: $\frac{x-1}{1} = \frac{y-0}{-1} = \frac{z+1}{-3} = \mu$.
Intersection with yz-plane (Put $x=0$):
$$\mu = -1$$
$$y = -(-1) = 1, \quad z = -1 – 3(-1) = 2$$
So, $R(0, 1, 2)$.
We need area of parallelogram PQRS. Area $= |\vec{PQ} \times \vec{PS}|$.
$\vec{PQ} = Q – P = (0-1)\hat{i} + (2-1)\hat{j} + (4-1)\hat{k} = -\hat{i} + \hat{j} + 3\hat{k}$.
$\vec{PS} = S – P = (1-1)\hat{i} + (0-1)\hat{j} + (-1-1)\hat{k} = 0\hat{i} – \hat{j} – 2\hat{k}$.
Cross product:
$$\vec{PQ} \times \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 3 \\ 0 & -1 & -2 \end{vmatrix}$$
$$= \hat{i}(-2 – (-3)) – \hat{j}(2 – 0) + \hat{k}(1 – 0)$$
$$= \hat{i} – 2\hat{j} + \hat{k}$$
Area $= \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
Square of Area $= 6$.
Ans. (6)
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