Question

Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i}  - \hat{j} + 2 \hat{k}  \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j}  + 2 \hat{k} \right) = 6\]

 

Answer 1

We know that the equation of line passing through (1,2,3) is given by

`(x - 1)/a_1 = (y - 2)/b_1 = (z - 3)/c_1`      ..........(1)

We know that line `(x - x_1)/a_1 = (y - y_1)/b_1 = (z - z_1)/c_1` is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(2)

Here, line (1) is parallel to plane,

x – y + 2z = 5

So,

a × 1 + b × – 1 + c × 2 = 0

⇒ a – b + 2c = 0 ……(3)

Also, line (1) is parallel to plane,

3x + y + z = 6

So,

a × 3 + b × 1 + c × 1 = 0

⇒ 3a + b + c = 0 ……(4)

Solving equation (3) and (4) by cross multiplication we have,

`a/(-1 xx 1 - (1) xx 2) = b/(3 xx 2 - 1 xx 1) = c/(1 xx 1 - 3 xx -1)`

⇒ `a/(-1 - 2) = b/(6 - 1) = c/(1 + 3)`

⇒ `a/-3 = b/5 = c/4 = "k"("let")`

∴ a = – 3k, b = 5k and c = 4k

Putting the value in equation (1)

`(x - 1)/(-3k) = (y - 2)/(5k) = (z - 3)/(4k)`

Multiplying by k we have

`(x - 1)/(-3) = (y - 2)/(5) = (z - 3)/(4)`

The required equation is

`(x - 1)/(-3) = (y - 2)/(5) = (z - 3)/(4)`

`vecr = (hat1 + 2hatj + 3hatk) + k(-3hati + 5hatj + 4hatk)`