Name: Angle Between a Line and a Plane
Uploaded: 2019-07-19
Duration: 3 min 24 s
Description: While writing, different people hold the pen at different angles to the plane of the paper. Here is a method to find the angle between a plane and a line int...

Question

Find the Cartesian and vector equations of the planes passing through the intersection of the planes

\vec{r} \cdot (i + 3 j) + 6 = 0

and

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

, whose perpendicular distance from the origin is

1 unit

.

Accepted Answer

Given that, the equation of the planes are $\vec{r} \cdot (i + 3 j) + 6 = 0$ and $\vec{r} \cdot (3 i - j + 4 k) = 0$ .

Also, given that, the required plane is passing through the intersection of the given planes.

$∴$ Equation of required plane is $(x + 3 y + 6) + λ (3 x - y + 4 z) = 0$

$\Rightarrow (1 + 3 λ) x + (3 - λ) y + 4 λ z + 6 = 0$

Perpendicular distance to the plane from origin is unity.

Therefore, $\frac{6}{\sqrt{{(1 + 3 λ)}^{2} + {(3 - λ)}^{2} + {(4 λ)}^{2}}} = 1$

Since, perpendicular distance to a plane $a x + b y + c z + d = 0$ from a point $(x_{1}, y_{1}, z_{1})$ is $\frac{|a x_{1} + b y_{1} + c z_{1} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}}$ .

$\Rightarrow 36 = 1 + 9 λ^{2} + 6 λ + 9 + λ^{2} - 6 λ + 16 λ^{2}$

$\Rightarrow λ = \pm 1$

Equations of required planes are $(x + 3 y + 6) + (1) (3 x - y + 4 z) = 0$ and $(x + 3 y + 6) + (- 1) (3 x - y + 4 z) = 0$

$\Rightarrow 4 x + 2 y + 4 z + 6 = 0$ and $- 2 x + 4 y - 4 z + 6 = 0$

$\Rightarrow 2 x + y + 2 z + 3 = 0$ and $x - 2 y - 2 z - 3 = 0$

Hence, the required equation of the planes are $2 x + y + 2 z + 3 = 0$ and $x - 2 y + 2 z - 3 = 0$ .

Find the Cartesian and vector equations of the planes passing through the intersection of the planes $\vec{r} \cdot (i + 3 j) + 6 = 0$ and $\vec{r} \cdot (3 i - j + 4 k) = 0$ , whose perpendicular distance from the origin is $1 unit$ .

Important Questions on The Plane

Learn and Prepare for any exam you want !

Find the Cartesian and vector equations of the planes passing through the intersection of the planes r→·(i+3j)+6=0 and r→·(3i-j+4k)=0, whose perpendicular distance from the origin is 1 unit.

Important Questions on The Plane

Learn and Prepare for any exam you want !

Find the Cartesian and vector equations of the planes passing through the intersection of the planes $\vec{r} \cdot (i + 3 j) + 6 = 0$ and $\vec{r} \cdot (3 i - j + 4 k) = 0$ , whose perpendicular distance from the origin is $1 unit$ .