If the line \(\cfrac{\text x-3}2=\cfrac{y+2}{-1}=\cfrac{z+4}3\) lies i

1.	If the line \(\cfrac{\text x-3}2=\cfrac{y+2}{-1}=\cfrac{z+4}3\) lies in the plane lx + my - z = 9, then find the value of l2 + m2?
Answer» We know that the lines \(\cfrac{\text x-\text x_1}{l_1}=\cfrac{y-y_1}{m_1}=\cfrac{z-z_1}{n_1}\) lies in plane ax + by + cz + d = 0, then ax₁ + by₁ + cz₁ + d = 0 and al + bm + cn = 0 Here, x₁ = 3, y₁ = –2, z₁ = –4 and l = 2, m = –1, n = 3 a = l, b = m, c = –1, d = –9 i.e, 3l + (– 2)m + (– 4)(– 1) – 9 = 0 and 2l – m – 3 = 0 3l – 2m = 5 and 2l – m = 3 3l – 2m = 5 …… (1) 2l – m = 3 ……(2) Multiply eq.(1) by 2 and eq.(2) by 3 and then subtract we get m = –1 l = 1 l² + m² = 2

If the line \(\cfrac{\text x-3}2=\cfrac{y+2}{-1}=\cfrac{z+4}3\) lies in the plane lx + my - z = 9, then find the value of l2 + m2?

Answer»

We know that the lines \(\cfrac{\text x-\text x_1}{l_1}=\cfrac{y-y_1}{m_1}=\cfrac{z-z_1}{n_1}\) lies in plane ax + by + cz + d = 0, then

ax₁ + by₁ + cz₁ + d = 0 and al + bm + cn = 0

Here,

x₁ = 3, y₁ = –2, z₁ = –4 and l = 2, m = –1, n = 3

a = l, b = m, c = –1, d = –9

i.e, 3l + (– 2)m + (– 4)(– 1) – 9 = 0 and 2l – m – 3 = 0

3l – 2m = 5 and 2l – m = 3

3l – 2m = 5 …… (1)

2l – m = 3 ……(2)

Multiply eq.(1) by 2 and eq.(2) by 3 and then subtract we get

m = –1

l = 1

l² + m² = 2