Let A and B be two distinct points on the lin | vedclass

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(B) Let the line $L$ be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point on the line is $Q(3\lambda+6, 2\lambda+7, -2\lambda+7)$.

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(B) Let the line $L$ be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point on the line is $Q(3\lambda+6, 2\lambda+7, -2\lambda+7)$.Let $P = (1, 2, 3)$. The vector $\overrightarrow{PQ} = (3\lambda+5, 2\lambda+5, -2\lambda+4)$.Since $PQ$ is perpendicular to the line $L$ with direction vector $\vec{b} = (3, 2, -2)$,we have $\overrightarrow{PQ} \cdot \vec{b} = 0$.$3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$$17\lambda + 17 = 0 \Rightarrow \lambda = -1$.Thus,the foot of the perpendicular is $Q(3(-1)+6, 2(-1)+7, -2(-1)+7) = Q(3, 5, 9)$.Points $A$ and $B$ are at a distance $d = 2\sqrt{17}$ from $Q$ on the line. Let the unit vector along the line be $\hat{u} = \frac{1}{\sqrt{3^2+2^2+(-2)^2}}(3, 2, -2) = \frac{1}{\sqrt{17}}(3, 2, -2)$.The points $A$ and $B$ are given by $Q \pm d\hat{u} = (3, 5, 9) \pm 2\sqrt{17} \cdot \frac{1}{\sqrt{17}}(3, 2, -2) = (3, 5, 9) \pm 2(3, 2, -2)$.$A = (3+6, 5+4, 9-4) = (9, 9, 5)$ and $B = (3-6, 5-4, 9+4) = (-3, 1, 13)$.Then $\overrightarrow{OA} \cdot \overrightarrow{OB} = (9, 9, 5) \cdot (-3, 1, 13) = (9)(-3) + (9)(1) + (5)(13) = -27 + 9 + 65 = 47$.

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