**Line
of Intersection of Two Planes**

**Line of Intersection of Two Planes: **Let two non-parallel planes be \(\vec{r}.{{\vec{n}}_{1}}={{d}_{1}}\)
and \(\vec{r}.{{\vec{n}}_{2}}={{d}_{2}}\) now line of
intersection of planes is perpendicular to vectors \({{\vec{n}}_{1}}\) and \({{\vec{n}}_{2}}\). Therefore, line of intersection is parallel to vector \({{\vec{n}}_{1}}\times
{{\vec{n}}_{2}}\).

find the equation of the of intersection of plane \({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z-{{d}_{1}}=0\) and \({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z-{{d}_{2}}=0\), then we find any point on the line by putting z = 0(say), then we can find corresponding values of x and y by solving equations \({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z-{{d}_{1}}=0\) and \({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z-{{d}_{2}}=0\). Thus, by fixing the value of z = λ, we can find the corresponding value of x and y in terms of λ.

**Example: **Reduce
the equation of line x – y + 2z = 5 and 3x + y + z = 6 in symmetrical form.

**Solution:**

Given x – y + 2z = 5 and 3x + y + z = 6

Let z = λ

x – y + 2λ = 5…(1) and

3x + y + λ = 6…(2)

Solving equation (1) and (2)

x – y + 2λ = 5

3x + y + λ = 6

______________

4x + 3λ = 11

______________

4x = 11 – 3λ

And

From equation (1)

x – y + 2λ = 5

4x – 4y + 8λ = 20

(11 -3λ) – 4y + 8λ = 20

4y = 5λ – 9

4x = 11 -3λ and 4y = 5λ – 9

The equation of the line is \(\frac{4x-11}{-3}=\frac{4y+9}{5}=\frac{z-0}{1}\)