Name: Distance of a Point From a Plane
Uploaded: 2019-07-19
Duration: 9 min 54 s
Description: Three-dimensional space has an infinite number of points and planes in it! This video discusses the distance of a point from a plane, both in vector and cart...

Question

Find the vector equation of the plane passing through three points with position vectors

\hat{i} + \hat{j} - 2 \hat{k}, 2 \hat{i} - \hat{j} + \hat{k}

and

\hat{i} + 2 \hat{j} + \hat{k} .

Also, find the coordinates of the point of intersection of this plane and the line

\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + λ (2 \hat{i} - 2 \hat{j} + \hat{k}) .

Accepted Answer

Let $A (1, 1 , -2), B (2, -1, 1)$ and $C (1, 2, 1)$ be the points represented by the given position vectors.
The required plane passes through the point $A (1, 1, - 2)$ whose position vectors $\vec{a} = \hat{i} + \hat{j} - 2$ and is normal to the vector $\vec{n}$ given by $\vec{n} = \vec{A B} \times \vec{A C .}$

Clearly, $\vec{A B} = \vec{O B} - \vec{O A} = (2 \hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2 \hat{k}) = \hat{i} - 2 \hat{j} + 3 \hat{k}$
Since, $\vec{A B} = Position vector of \vec{B} - Position vector of \vec{A}$ .
$\vec{A C} = \vec{O C} - \vec{O A} = (\hat{i} + 2 \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2 \hat{k}) = 0 \hat{i} + \hat{j} + 3 \hat{k}$

$\vec{n} = \vec{A B} \times \vec{A C} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 0 & 1 & 3 \end{matrix}| = - 9 \hat{i} - 3 \hat{j} + \hat{k}$

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

$\Rightarrow \vec{r} \cdot (- 9 \hat{i} - 3 \hat{j} + \hat{k}) = (\hat{i} + \hat{j} - 2 \hat{k}) \cdot (- 9 \hat{i} - 3 \hat{j} + \hat{k})$

$\Rightarrow \vec{r} \cdot (- 9 \hat{i} - 3 \hat{j} + \hat{k}) = - 9 - 3 - 2$

$\Rightarrow \vec{r} \cdot [- (9 \hat{i} + 3 \hat{j} - \hat{k})] = - 14$

$\Rightarrow \vec{r} \cdot (9 \hat{i} + 3 \hat{j} - \hat{k}) = 14$ ....(i)

To find the point of intersection of this plane, the given equation of the line is $\vec{r} = (3 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 2 \hat{j} + \hat{k})$

$\Rightarrow \vec{r} = (3 + 2 λ) \hat{i} + (- 1 - 2 λ) \hat{j} + (- 1 + λ) \hat{k}$

The coordinates of any point on this line are in the form of $(3 + 2 λ), (- 1 - 2 λ), (- 1 + λ)$ .

Since, this point lies on the plane $\vec{r} \cdot (9 \hat{i} + 3 \hat{j} - \hat{k}) = 14$ (from equation(i))

$[(3 + 2 λ) \hat{i} + (- 1 - 2 λ) \hat{j} + (- 1 + λ) \hat{k}] . (9 \hat{i} + 3 \hat{j} - \hat{k}) = 14$

$\Rightarrow 27 + 18 λ - 3 - 6 λ + 1 - λ = 14$

$\Rightarrow 11 λ = - 11$

$\Rightarrow λ = - 1$

So, the coordinates of the point are $(3 + 2 λ, - 1 - 2 λ, - 1 + λ)$

$= (3 - 2, - 1 + 2, - 1 - 1)$

$= (1, 1, - 2)$ .

Important Questions on The Plane

Learn and Prepare for any exam you want !

Find the vector equation of the plane passing through three points with position vectors i^+j^-2k^, 2i^-j^+k^ and i^+2j^+k^. Also, find the coordinates of the point of intersection of this plane and the line r→=3i^-j^-k^+λ2i^-2j^+k^.

Important Questions on The Plane

Learn and Prepare for any exam you want !