Question

Find vector equation of a plane which is at a distance of 42 units from origin and which is normal to the vector 2i+j-2k​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \bull \: \: \sf \: Plane \:at \: a \: distance \: of \: 42 \: units \: from \: origin.$

$\: \: \: \: \bull \: \: \sf \: Normal \: vector \: to \: the \: plane \: 2 \: \hat{i} + \hat{j} - 2 \hat{k}$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \bull \: \sf \: \: Vector \: equation \: of \: plane$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\sf \: Unit \: Vector \: in \: the \: direction \: of \: \vec{a} = \dfrac{\vec{a}}{ |\vec{a}| }$

Equation of plane which is at a distance of d units from the

$\sf \: origin \: and \: normal \: to \: unit \: vector \: \hat{n} \: is \: given \: by \:$

$\sf \: \vec{r} \: . \: \hat{n} \: = \: d$

$\large\underline{\sf{Solution-}}$

Here,

$\sf \: Normal \: vector \: to \: the \: plane \: 2 \: \hat{i} + \hat{j} - 2 \hat{k}$

So,

$\rm :\longmapsto\: |\vec{n}| = \sqrt{ {2}^{2} + {1}^{2} + {( - 2)}^{2} }$

$\rm :\longmapsto\: |\vec{n}| = \sqrt{2 + 1 + 4}$

$\rm :\longmapsto\: |\vec{n}| = \sqrt{9}$

$\rm :\implies\: |\vec{n}| = 3$

$\sf \: \therefore \: \: \hat{n} = \dfrac{\vec{n}}{ |\vec{n}| }$

$\rm :\longmapsto\:\hat{n} = \dfrac{2\hat{i} + \hat{j} - 2\hat{k}}{3}$

Now,

It is given that

• Plane is at a distance of 42 units from the origin,

$\rm :\implies\:d \: = \: 42$

Now,

Equation of plane is given by

$\rm :\longmapsto\: \: \vec{r} \: . \: \hat{n} \: = \: d$

$\rm :\longmapsto\:\vec{r}.\bigg(\dfrac{2\hat{i} + \hat{j} - 2\hat{k}}{3}\bigg) = 42$

$\rm :\implies\:\vec{r}.(2\hat{i} + \hat{j} - 2\hat{k}) = 126$

Additional Information :-

• 1. The general equation of plane is given by ax + by + cz + d = 0.

• 2. The scalar components of normal to plane is (a, b, c).

3. The equation of plane passes through the point (p, q, r) havimg direction ratios (a, b, c) respectively, is a(x - p) + b(y - q) + c(z - r) = 0