P and Q are points on the straight line passi | vedclass

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(A) The line passes through $A(3, 1, -1)$ and is parallel to $\vec{v} = 2 \hat{i} - \hat{j} + 2 \hat{k}$. The unit vector along the line is $\hat{u} =

Vedclass · Accepted Answer

$r=(s+5t) \hat{i} + 2s \hat{j} + (t-3s) \hat{k}$

Vedclass · Answer

(A) The line passes through $A(3, 1, -1)$ and is parallel to $\vec{v} = 2 \hat{i} - \hat{j} + 2 \hat{k}$. The unit vector along the line is $\hat{u} = \frac{2 \hat{i} - \hat{j} + 2 \hat{k}}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{2 \hat{i} - \hat{j} + 2 \hat{k}}{3}$.Since $AP = AQ = 3$,the points $P$ and $Q$ are given by $\vec{A} \pm 3\hat{u}$.$P, Q = (3 \hat{i} + \hat{j} - \hat{k}) \pm 3 \left( \frac{2 \hat{i} - \hat{j} + 2 \hat{k}}{3} \right) = (3 \hat{i} + \hat{j} - \hat{k}) \pm (2 \hat{i} - \hat{j} + 2 \hat{k})$.Thus,$P = (3+2) \hat{i} + (1-1) \hat{j} + (-1+2) \hat{k} = 5 \hat{i} + \hat{k}$ and $Q = (3-2) \hat{i} + (1+1) \hat{j} + (-1-2) \hat{k} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.The plane $OPQ$ passes through the origin $O(0,0,0)$ and contains vectors $\vec{OP} = 5 \hat{i} + \hat{k}$ and $\vec{OQ} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.The vector equation of the plane is $\vec{r} = s \vec{OP} + t \vec{OQ} = s(5 \hat{i} + \hat{k}) + t(\hat{i} + 2 \hat{j} - 3 \hat{k}) = (5s+t) \hat{i} + 2t \hat{j} + (s-3t) \hat{k}$.Comparing with the options,option $A$ represents the same plane by re-parameterizing $s$ and $t$.

$P$ and $Q$ are points on the straight line passing through the point $A(3 \hat{i}+\hat{j}-\hat{k})$ and parallel to the vector $\vec{v} = 2 \hat{i}-\hat{j}+2 \hat{k}$. If $AP = AQ = 3$,then the vector equation of the plane $OPQ$ is:

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