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(C) The normal vector $\vec{n}$ to the plane is parallel to the line segment $\vec{AB}$.
$\vec{AB} = (-1 - (-2))\hat{i} + (-16 - (-21))\hat{j} + (33

Vedclass · Accepted Answer

$\frac{188}{121}$

Vedclass · Answer

(C) The normal vector $\vec{n}$ to the plane is parallel to the line segment $\vec{AB}$.
$\vec{AB} = (-1 - (-2))\hat{i} + (-16 - (-21))\hat{j} + (33 - 29)\hat{k} = 1\hat{i} + 5\hat{j} + 4\hat{k}$.
The plane passes through $Q(4, -2, 2)$ and $P(\lambda, 2, 1)$. Thus,the vector $\vec{PQ}$ lies in the plane.
$\vec{PQ} = (4 - \lambda)\hat{i} + (-2 - 2)\hat{j} + (2 - 1)\hat{k} = (4 - \lambda)\hat{i} - 4\hat{j} + 1\hat{k}$.
Since the plane is perpendicular to the line $AB$,the normal vector $\vec{AB}$ is perpendicular to any vector in the plane,including $\vec{PQ}$.
Therefore,$\vec{AB} \cdot \vec{PQ} = 0$.
$(1\hat{i} + 5\hat{j} + 4\hat{k}) \cdot ((4 - \lambda)\hat{i} - 4\hat{j} + 1\hat{k}) = 0$.
$1(4 - \lambda) + 5(-4) + 4(1) = 0$.
$4 - \lambda - 20 + 4 = 0$.
$-\lambda - 12 = 0 \Rightarrow \lambda = -12$.
Now,calculate the expression $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.
Substitute $\lambda = -12$: $\left(\frac{-12}{11}\right)^{2} - \frac{4(-12)}{11} - 4 = \frac{144}{121} + \frac{48}{11} - 4 = \frac{144 + 528 - 484}{121} = \frac{188}{121}$.

Let $P(\lambda, 2, 1)$ be a point on the plane which passes through the point $Q(4, -2, 2)$. If the plane is perpendicular to the line joining the points $A(-2, -21, 29)$ and $B(-1, -16, 33)$,then find the value of $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.

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