The vector equation of a plane in parametric | vedclass

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Question

(A) Step $1$: Identify the position vector of a point on the plane. Let $\vec{a} = -\hat{i} + 2\hat{j}$.Step $2$: Find the first direction vector $\ve

Vedclass · Accepted Answer

$\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$

Vedclass · Answer

(A) Step $1$: Identify the position vector of a point on the plane. Let $\vec{a} = -\hat{i} + 2\hat{j}$.Step $2$: Find the first direction vector $\vec{b}$ by the vector joining points $A(-1, 2, 0)$ and $B(2, 2, -1)$. $\vec{b} = (2 - (-1))\hat{i} + (2 - 2)\hat{j} + (-1 - 0)\hat{k} = 3\hat{i} - \hat{k}$.Step $3$: Find the second direction vector $\vec{c}$ from the line $\frac{x - 1}{1} = \frac{y + 1/2}{1} = \frac{z + 1}{-1}$. The direction ratios are $(1, 1, -1)$, so $\vec{c} = \hat{i} + \hat{j} - \hat{k}$.Step $4$: The parametric form of the plane is $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.Substituting the vectors: $\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$.This matches option $A$.

The vector equation of a plane in parametric form, passing through the points $A(-1, 2, 0)$ and $B(2, 2, -1)$ and parallel to the line $\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{-1}$ is:

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