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 equation of the plane which contains the line of intersection of the planes

\vec{r} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0, \vec{r} \cdot (2 \hat{i} + \hat{j} - \hat{k}) + 5 = 0

and which is perpendicular to the plane

\vec{r} \cdot (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0

.

Accepted Answer

The Cartesian equations of the planes $\vec{r} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0$ and $\vec{r} \cdot (2 \hat{i} + \hat{j} - \hat{k}) + 5 = 0$ can be obtained by substituting $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ in the above equations. Therefore, we get:

$(x \hat{i} + y \hat{j} + z \hat{k}) . (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0 \Rightarrow x + 2 y + 3 z - 4 = 0 \dots (1)$

$(x \hat{i} + y \hat{j} + z \hat{k}) . (2 \hat{i} + \hat{j} - \hat{k}) + 5 = 0 \Rightarrow 2 x + y - z + 5 = 0 \dots (2)$

The equation of a plane passing through the intersection of two planes is given by $P_{1} + λ P_{2} = 0$ . Therefore, the equations $(1)$ and $(2)$ can be related as below:

$(x + 2 y + 3 z - 4) + λ (2 x + y - z + 5) = 0$

Rearranging the terms in the above equation, we get:

$(1 + 2 λ) x + (2 + λ) y + (3 - λ) z + (- 4 + 5 λ) = 0 \dots (3)$

According to the question, the above plane is perpendicular to the plane $\vec{r} \cdot (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$ . We know that, the direction ratios of this plane are $(5, 3, - 6)$ . So the condition for perpendicular planes can be applied as below by relating equation $(3)$ and the direction ratios of the above plane, we get:

$(1 + 2 λ) 5 + (2 + λ) 3 + (3 - λ) \times - 6 = 0$

$5 + 10 λ + 6 + 3 λ - 18 + 6 λ = 0$

$19 λ - 7 = 0 \Rightarrow λ = \frac{7}{19}$

Substituting the value of $λ$ in equation $(3)$ , we get: $(1 + 2 (\frac{7}{19})) x + (2 + \frac{7}{19}) y + (3 - \frac{7}{19}) z + (- 4 + 5 (\frac{7}{19})) = 0$

$\frac{33}{19} x + \frac{45}{19} y + \frac{50}{19} z - \frac{41}{19} = 0$

$\Rightarrow 33 x + 45 y + 50 z - 41 = 0$

Hence, the equation of the plane is $33 x + 45 y + 50 z - 41 = 0$ .

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