The equation of a plane containing the line x | vedclass

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(D) The direction ratios of the given lines are $\vec{b_1} = (1, 4, 7)$ and $\vec{b_2} = (3, 5, 7)$.Since the plane contains the first line and is par

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$x-2y+z=0$

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(D) The direction ratios of the given lines are $\vec{b_1} = (1, 4, 7)$ and $\vec{b_2} = (3, 5, 7)$.Since the plane contains the first line and is parallel to the second,the normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$.$\vec{n} = \vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 4 & 7 \ 3 & 5 & 7 \end{vmatrix}$$= \hat{i}(28-35) - \hat{j}(7-21) + \hat{k}(5-12) = -7\hat{i} + 14\hat{j} - 7\hat{k} = -7(\hat{i} - 2\hat{j} + \hat{k})$.Thus,the equation of the plane is of the form $x - 2y + z = d$.Since the plane contains the line,it must pass through the point $(2, 4, 6)$.Substituting the point into the equation: $2 - 2(4) + 6 = 2 - 8 + 6 = 0$.Therefore,the equation of the plane is $x - 2y + z = 0$.

The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the line $\vec{r}=(\hat{i}+3\hat{j}+5\hat{k})+\lambda(3\hat{i}+5\hat{j}+7\hat{k})$ is

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