Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The normal vector $\vec{n}$ to the plane is perpendicular to both the given vectors. Thus, it is given by their cross product:

$\vec{n}=(4\hat{i}-\hat{j}+3\hat{k})\times (\hat{i}+\hat{j}+\hat{k})$

$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 3\\ 1& 1& 1\end{array}\right|$

$\vec{n}=\hat{i}(-1-3)-\hat{j}(4-3)+\hat{k}(4+1)$

$\vec{n}=-4\hat{i}-\hat{j}+5\hat{k}$

The plane passes through the point $A(-2,7,5)$, so its position vector is $\vec{a}=-2\hat{i}+7\hat{j}+5\hat{k}$.

The vector equation of the plane is given by $\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$.

Calculating $\vec{a}\cdot \vec{n}$:

$\vec{a}\cdot \vec{n}=(-2\hat{i}+7\hat{j}+5\hat{k})\cdot (-4\hat{i}-\hat{j}+5\hat{k})$

$\vec{a}\cdot \vec{n}=(-2)(-4)+(7)(-1)+(5)(5)$

$\vec{a}\cdot \vec{n}=8-7+25=26$

Substituting these into the equation of the plane, we get:

$\vec{r}\cdot (-4\hat{i}-\hat{j}+5\hat{k})=26$

Answer: $\bar{r}\cdot (-4\hat{i}-\hat{j}+5\hat{k})=26$