The plane 3x + 4y + 6z + 7 = 0 is rotated abo | vedclass

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Question

(A) The equation of a plane passing through the origin and containing the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})

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$x + y + z = 0$

Vedclass · Answer

(A) The equation of a plane passing through the origin and containing the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ is given by the scalar triple product of the position vector of a point on the line,the direction vector of the line,and the general vector $r = x\hat{i} + y\hat{j} + z\hat{k}$.Since the plane passes through the origin $(0, 0, 0)$,the equation is $(r - 0) \cdot [(\hat{i} + 2\hat{j} - 3\hat{k}) 	imes (2\hat{i} - 3\hat{j} + \hat{k})] = 0$.Calculating the cross product:$(\hat{i} + 2\hat{j} - 3\hat{k}) 	imes (2\hat{i} - 3\hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & -3 & 1 \end{vmatrix} = \hat{i}(2 - 9) - \hat{j}(1 + 6) + \hat{k}(-3 - 4) = -7\hat{i} - 7\hat{j} - 7\hat{k}$.Thus,the equation of the plane is $r \cdot (-7\hat{i} - 7\hat{j} - 7\hat{k}) = 0$,which simplifies to $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.This corresponds to the Cartesian equation $x + y + z = 0$.Therefore,option $A$ is correct.

The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is

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