The vector equation of the plane containing t | vedclass

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Question

(C) The two given lines pass through the point with position vector $a=\hat{i}+\hat{j}$ and are parallel to the vectors $b_1=\hat{i}+2 \hat{j}-\hat{k}

Vedclass · Accepted Answer

$r \cdot n=0$, where $n=\hat{i}-\hat{j}-\hat{k}$

Vedclass · Answer

(C) The two given lines pass through the point with position vector $a=\hat{i}+\hat{j}$ and are parallel to the vectors $b_1=\hat{i}+2 \hat{j}-\hat{k}$ and $b_2=-\hat{i}+\hat{j}-2 \hat{k}$ respectively. The plane containing these lines passes through the point $a=\hat{i}+\hat{j}$ and is normal to the vector $n = b_1 	imes b_2$.Calculating the normal vector $n$:$n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ -1 & 1 & -2 \end{vmatrix} = \hat{i}(-4+1) - \hat{j}(-2-1) + \hat{k}(1+2) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.The vector equation of the plane is $r \cdot n = a \cdot n$.$r \cdot (-3\hat{i} + 3\hat{j} + 3\hat{k}) = (\hat{i} + \hat{j}) \cdot (-3\hat{i} + 3\hat{j} + 3\hat{k}) = -3 + 3 + 0 = 0$.Dividing by $-3$, we get $r \cdot (\hat{i} - \hat{j} - \hat{k}) = 0$.

The vector equation of the plane containing the lines $r=(\hat{i}+\hat{j})+t(\hat{i}+2 \hat{j}-\hat{k})$ and $r=(\hat{i}+\hat{j})+s(-\hat{i}+\hat{j}-2 \hat{k})$ is

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