The distance of the origin from the plane r c | vedclass

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Question

(B) The equation of the line passing through the origin and parallel to the vector $\vec{v} = 6 \hat{i}+2 \hat{j}+3 \hat{k}$ is given by $\frac{x}{6}

Vedclass · Accepted Answer

$\frac{49}{10}$

Vedclass · Answer

(B) The equation of the line passing through the origin and parallel to the vector $\vec{v} = 6 \hat{i}+2 \hat{j}+3 \hat{k}$ is given by $\frac{x}{6} = \frac{y}{2} = \frac{z}{3} = k$. Any point $P$ on this line is of the form $(6k, 2k, 3k)$. Since this point $P$ lies on the plane $3x + 4y - 12z = 7$, we substitute the coordinates of $P$ into the plane equation: $3(6k) + 4(2k) - 12(3k) = 7$ $18k + 8k - 36k = 7$ $-10k = 7 \Rightarrow k = -\frac{7}{10}$. The coordinates of point $P$ are $(6(-\frac{7}{10}), 2(-\frac{7}{10}), 3(-\frac{7}{10})) = (-\frac{42}{10}, -\frac{14}{10}, -\frac{21}{10})$. The distance from the origin $(0, 0, 0)$ to point $P$ is given by the distance formula: $d = \sqrt{(-\frac{42}{10})^2 + (-\frac{14}{10})^2 + (-\frac{21}{10})^2}$ $d = \frac{1}{10} \sqrt{42^2 + 14^2 + 21^2} = \frac{1}{10} \sqrt{1764 + 196 + 441} = \frac{1}{10} \sqrt{2401} = \frac{49}{10}$.

The distance of the origin from the plane $r \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})=7$ measured parallel to the line $r=(\hat{i}+2 \hat{j}+3 \hat{k})+t(6 \hat{i}+2 \hat{j}+3 \hat{k})$ is

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