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(D) The equation of a plane passing through the origin $(0, 0, 0)$ and containing the $y$-axis (which has direction vector $\hat{j}$) and passing thro

Vedclass · Accepted Answer

$3x - z = 0$

Vedclass · Answer

(D) The equation of a plane passing through the origin $(0, 0, 0)$ and containing the $y$-axis (which has direction vector $\hat{j}$) and passing through the point $P(1, 2, 3)$ can be found using the normal vector $\vec{n}$.The plane contains the vector $\vec{v} = \hat{j} = (0, 1, 0)$ and the vector $\vec{OP} = \hat{i} + 2\hat{j} + 3\hat{k} = (1, 2, 3)$.The normal vector $\vec{n}$ is given by the cross product:$\vec{n} = \vec{v} 	imes \vec{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 0 \ 1 & 2 & 3 \end{vmatrix}$$\vec{n} = \hat{i}(3 - 0) - \hat{j}(0 - 0) + \hat{k}(0 - 1) = 3\hat{i} - \hat{k}$.The equation of the plane passing through $(0, 0, 0)$ with normal $\vec{n} = (3, 0, -1)$ is:$3(x - 0) + 0(y - 0) - 1(z - 0) = 0$$3x - z = 0$.

The equation of the plane which contains the $y$-axis and passes through the point $(1, 2, 3)$ is:

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