The equation of a plane containing the lines | vedclass

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(D) The normal vector $\vec{n}$ of the plane is perpendicular to the direction vectors of both lines,$\vec{v}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and

Vedclass · Accepted Answer

$9 x-8 y+z+11=0$

Vedclass · Answer

(D) The normal vector $\vec{n}$ of the plane is perpendicular to the direction vectors of both lines,$\vec{v}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $\vec{v}_2 = \hat{i} + \hat{j} - \hat{k}$.Thus,$\vec{n} = \vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 6 \ 1 & 1 & -1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(-2-6) + \hat{k}(2-3) = -9\hat{i} + 8\hat{j} - \hat{k}$.We can take the normal vector as $\vec{n} = 9\hat{i} - 8\hat{j} + \hat{k}$.The plane passes through the point $(1, 3, 4)$ from the second line equation.The equation of the plane is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,where $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$.$(x\hat{i} + y\hat{j} + z\hat{k} - (\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (9\hat{i} - 8\hat{j} + \hat{k}) = 0$.$9(x-1) - 8(y-3) + 1(z-4) = 0$.$9x - 9 - 8y + 24 + z - 4 = 0$.$9x - 8y + z + 11 = 0$.

The equation of a plane containing the lines $\overline{r}=(\hat{\imath}+2 \hat{\jmath}-4 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k})$ and $\overline{r}=(\hat{\imath}+3 \hat{\jmath}+4 \hat{k})+\mu(\hat{\imath}+\hat{\jmath}-\hat{k})$ is

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