Question

Find the vector equation of the following planes in scalar product form:
$straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space plus space straight mu space open parentheses straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses$

Solution

The equation of plane is

straight r with rightwards arrow on top space equals space straight i with hat on top space space minus space straight j with hat on top space plus space straight lambda open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space plus space straight mu space open parentheses straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses

straight x straight i with hat on top space space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top space equals space open parentheses 1 plus straight lambda space plus space straight mu close parentheses space straight i with hat on top space plus space open parentheses negative 1 plus straight lambda space minus space 2 space straight mu close parentheses straight j with hat on top space plus space open parentheses straight lambda plus space 3 space straight mu close parentheses space straight k with hat on top

Equating the coefficients of

straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top comma

we get,
x = 1 + λ + μ    ....(1)
y = – 1 + λ – 2 μ    ...(2)
z = λ + 3 μ    ...(3)
We are to eliminate λ and μ from (1), (2), (3).
Subtracting (2) from (1), we get,
x – y = 2 + 3 μ    ... (4)Subtracting (3) from (1), we get,
x – z = 1 – 2 μ    ... (5)
Multiplying (4) by 2 and (5) by 3 , we get,
2 x – 2 y = 4 + 6 μ    ...(6)
3 x – 3 z = 3 – 6 μ    ...(7)
Adding (6) and (7), we get, 5 x – 2 y – 3 z = 7
This can be written as

open parentheses straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top close parentheses. space open parentheses 5 space straight i with hat on top space space minus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top close parentheses space equals space 7

straight r with rightwards arrow on top. space open parentheses 5 space straight i with hat on top space minus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top close parentheses space equals space 7

which is the required equation.

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