Question

Find the vector equation of the plane in scalar product form
$straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda space open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space plus space straight mu space open parentheses 4 straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses.$

Solution

The equation of plane is

straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top right parenthesis space plus space straight mu space left parenthesis 4 space 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 right parenthesis

straight x straight i with hat on top space plus space straight y 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 plus 4 straight mu close parentheses straight i with hat on top space plus space left parenthesis negative 1 plus straight lambda minus 2 straight mu right parenthesis space straight j with hat on top space plus space left parenthesis straight lambda plus 3 straight mu right parenthesis 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 space space we space get comma

x = 1 + λ + 4 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 + 6 μ        ...(4)
Subtracting (3) from (1), we get,
x – z = 1 + n    ...(5)
Multiplying (4) by 1, (5) by –6, we get,
x – y = 2 + 6 μ    ...(6)
– 6 x + 6 z = – 6 – 6 μ    .....(7)
Adding (6) and (7), we get,
– 5 x – y + 6 z = – 4
or 5 x + y – 6 z = 4
or

open parentheses straight x 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 space open parentheses 5 space straight i with hat on top space plus space straight j with hat on top space minus space 6 space straight k with hat on top close parentheses space equals 4

straight r with rightwards arrow on top. space left parenthesis 5 straight i with hat on top space plus space straight j with hat on top space minus space 6 space straight k with hat on top right parenthesis space equals space 4 comma

which is required vector equation of plane.

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Three Dimensional Geometry

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