Name: The Shortest Distance Between Two Skew Lines
Uploaded: 2019-07-19
Duration: 9 min 21 s
Description: Skew lines are the lines that do not intersect and not parallel. The shortest distance between two skew lines lies along the line, which is perpendicular to...

Question

Find the vector equation of the plane passing through the intersection of the planes

\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7, \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9

and through the point

(2, 1, 3)

.

Accepted Answer

It is given that the plane passes through the intersection of the planes $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7$ and $\vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3)$ .

The equations of the planes are

$\Rightarrow \vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) - 7 = 0$ …(i)

$\vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9 = 0$ …(ii)

So, the equation of any plane through the intersection of the planes given in equations (i) and (ii) is given by,

$[\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) - 7] + λ [\vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9] = 0,$ where $λ \in R$

$\vec{r} \cdot [(2 \hat{i} + 2 \hat{j} - 3 \hat{k}) + λ (2 \hat{i} + 5 \hat{j} + 3 \hat{k})] = 9 λ + 7$

$\vec{r} \cdot [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7$ …(iii)

The plane passes through the point $(2, 1, 3) .$

Therefore, its position vector is given by,

$\vec{r} = 2 \hat{i} + \hat{j} + 3 \hat{k}$

By substituting in equation (iii), we obtain

$(2 \hat{i} + \hat{j} + 3 \hat{k}) \cdot [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7$

$\Rightarrow 2 (2 + 2 λ) + (2 + 5 λ) + 3 (3 λ - 3) = 9 λ + 7$

$\Rightarrow 18 λ - 3 = 9 λ + 7$

$\Rightarrow 9 λ = 10$

$\Rightarrow λ = \frac{10}{9}$
Hence, by substituting $λ = \frac{10}{9}$ in equation (iii), we obtain

$\vec{r} \cdot (\frac{38}{9} \hat{i} + \frac{68}{9} \hat{j} + \frac{3}{9} \hat{k}) = 17$

$\Rightarrow \vec{r} \cdot (38 \hat{i} + 68 \hat{j} + 3 \hat{k}) = 153$

Therefore, $\vec{r} \cdot (38 \hat{i} + 68 \hat{j} + 3 \hat{k}) = 153$ is the vector equation of the required plane.

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7, \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3)$ .

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the vector equation of the plane passing through the intersection of the planes r→·2i^+2j^-3k^=7,r→·2i^+5j^+3k^=9 and through the point 2,1,3.

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7, \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3)$ .