how cn we find point of intersection of line and plane ?

line given by : vectr r=2i-j+2k+lambda (3i+4j+2k) and plane given by : vector r.(i-j+k)=5

Question

how cn we find point of intersection of line and plane ?
line given by : vectr r=2i-j+2k+lambda (3i+4j+2k) and plane
given by : vector r (i-j+k)=5

Shridhar Kaushik · Accepted Answer

$Equation of given line is \vec{r} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) and Equation of given plane is : \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$

We have to find the point of intersection of line and plane.

Let P be the point of intersection of line and plane.

Eqⁿ of line in cartesian form can be written as:

$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{z} = λ (say)$

⇒ x = 3λ + 2, y = 4λ – 2, z = 2λ + 2

These are the coordinates of any general point which lies on the given line.Since P lies on the line, hence for same value of λ, coordinates of P will be (3λ + 2, 4λ – 2, 2λ + 2).

Also, Equation of plane in cartesian form is —

$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \Rightarrow x - y + z = 5$

Since point of intersection (P) also lies in the plane, so it will satisfies the equation of plane.

So, (3λ + 2) – (4λ – 2) + (2λ + 2) = 5

⇒ λ + 6 = 5

⇒ λ = – 1

Hence, coordinates of Point P are (–1, –6, 0).

Thus, the point of intersection of line and plane is (–1, –6, 0).

-6

how cn we find point of intersection of line and plane ? line given by : vectr r=2i-j+2k+lambda (3i+4j+2k) and plane given by : vector r.(i-j+k)=5

how cn we find point of intersection of line and plane ?

line given by : vectr r=2i-j+2k+lambda (3i+4j+2k) and plane given by : vector r.(i-j+k)=5