Let a line L pass through the point P(2, 3, 1 | vedclass

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(B) The direction vector $\vec{v}$ of the line is given by the cross product of the normals to the planes $x + 3y - 2z - 2 = 0$ and $x - y + 2z = 0$.$

Vedclass · Accepted Answer

$158$

Vedclass · Answer

(B) The direction vector $\vec{v}$ of the line is given by the cross product of the normals to the planes $x + 3y - 2z - 2 = 0$ and $x - y + 2z = 0$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & -2 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6 - 2) - \hat{j}(2 + 2) + \hat{k}(-1 - 3) = 4\hat{i} - 4\hat{j} - 4\hat{k}$.We can take the direction vector as $\vec{d} = (1, -1, -1)$.The equation of line $L$ passing through $P(2, 3, 1)$ is $\frac{x - 2}{1} = \frac{y - 3}{-1} = \frac{z - 1}{-1} = k$.Any point $R$ on the line is $(k + 2, -k + 3, -k + 1)$.Let $Q = (5, 3, 8)$. The vector $\vec{QR} = (k + 2 - 5, -k + 3 - 3, -k + 1 - 8) = (k - 3, -k, -k - 7)$.Since $\vec{QR}$ is perpendicular to the line direction $(1, -1, -1)$,we have:$1(k - 3) - 1(-k) - 1(-k - 7) = 0 \Rightarrow k - 3 + k + k + 7 = 0 \Rightarrow 3k + 4 = 0 \Rightarrow k = -\frac{4}{3}$.The vector $\vec{QR} = (-\frac{4}{3} - 3, -(-\frac{4}{3}), -(-\frac{4}{3}) - 7) = (-\frac{13}{3}, \frac{4}{3}, -\frac{17}{3})$.The distance $\alpha = |\vec{QR}| = \sqrt{(-\frac{13}{3})^2 + (\frac{4}{3})^2 + (-\frac{17}{3})^2} = \sqrt{\frac{169 + 16 + 289}{9}} = \sqrt{\frac{474}{9}}$.Thus,$\alpha^2 = \frac{474}{9}$.Therefore,$3\alpha^2 = 3 	imes \frac{474}{9} = \frac{474}{3} = 158$.

Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$,then $3\alpha^2$ is equal to $......$.

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