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QUESTION: 1

The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 is

Solution:

The equation of any plane passing through the line of intersection of two planes

As we have learnt from the concept Equation of plane

QUESTION: 2

The equation of the plane passing through the line of intersection of the planes x-2y+3z+8=0 and 2x-7y+4z-3=0 and the point (3, 1, -2) is:

Solution:

(x - 2y + 3z + 8) + μ(2x - 7y + 4z - 3) = 0

i.e, (1 + 2μ)x - (2 + 7μ)y + (3 + 4μ)z + (8 - 3μ) = 0......(1)

the required plane is passing through (3, 1, -2)

so, 3(1 + 2μ) - (1)(2 + 7μ) + (-2)(3 + 4μ)+ (8 - 3μ) = 0

3 + 6μ - 2 - 7μ -6 -8μ + 8 - 3μ = 0

by solving, μ = 1/4

putting μ in equation (1)

we get the required equation of plane as :- 6x - 15y + 16z + 29 = 0

QUESTION: 3

If are the equations of two given lines the, what is the condition for coplanarity of the twi lines in the vector form.

Solution:

QUESTION: 4

The equation of the plane which makes the intercepts 3, 4, 12 with X-axis, Y-axis and Z-axis respectively is:

Solution:

QUESTION: 5

The equation of the plane passing through the intersection of the planes and and the point (1, 2, 1) is:

Solution:

n1 = 2i + j + k

n2 = 2i + 3j - 4k

p1 = 4, p2 = -6

r.(n1 + λn2) = p1 + λp2

=> r . [2i + j + k + λ(2i + 3j - 4k)] = 4 - 6λ

=> r . [ i(2 + 2λ) + j(1 + 3λ) + k(1 - 4k)] = 4 - 6λ

Taking r = xi + yj + zk

(2 + 2λ)x + (1 + 3λ)y + (1 - 4k)z = 4 - 6λ

(2x + y + - z - 4) + λ(2x + 3y - 4k + 6) = 0

Given points are (1,2,1)

(2 + 2 - 1 - 4) + λ(2 + 6 - 4 + 6) = 0

-1 + λ(10) = 0

λ = 1/10

Substitute λ = 1/10, we get

18x + 7y + 14z - 46=0

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