Find equation of plane passing through (1,1,1) and containing the line r= (-3i+j+5k) +$(-3i-j-5k).

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Solution

Equation of plane

a(x - x1) + b(y - y1) + c(z - z1) = 0

given line

r = -3i + j + 5k + 3A i - Aj -5Ak

taking A = 1

r = 0, thus (0, 0, 0) is a point on the line

Take A = 2

r = 3i - j - 5k

(3, -1, -5) is also a point on the line

hence the plane passes through three points (0,0,0) , (1, 1, 1) and (3, -1,-5)

a(1 - 0) + b(1 - 0) + c(1 - 0) = 0

a + b + c = 0

a(3 - 0) + b(-1 - 0) + c(-5 - 0) = 0

3a - b - 5c = 0

a / (-5 + 1) = -b/(-5 - 3) = c/(-1 - 3)

a/-4 = b/8 = c/-4

(a :b:c) = (1: -2: 1)

hence the given plane is

ax + by + cz = 0

x - 2y + z = 0

is the required equation of the plane

its normal is

i - 2j + k

Suggest Corrections

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(- 2, 1, 3)

3 i + j + 5 k .

\vec{r} \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 5 and \vec{r} \cdot (3 \hat{i} + \hat{j} + \hat{k}) = 6 .

3 i + 4 j + 5 k, 4 i - 3 j - 5 k

7 i + j

\vec{r} \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 5 and \vec{r} \cdot (3 \hat{i} + \hat{j} + 2 \hat{k}) = 6

θ

\to r = 2 i + j - k + (i + j + k) t

\to r \cdot (3 i - 4 j + 5 k) = q

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