Test: Intercept Form And Family Of Planes - JEE MCQ

# Test: Intercept Form And Family Of Planes - JEE MCQ

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## 5 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Intercept Form And Family Of Planes

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Test: Intercept Form And Family Of Planes - Question 1

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

Detailed Solution for Test: Intercept Form And Family Of Planes - Question 1

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

As we have learnt from the concept Equation of plane

Test: Intercept Form And Family Of Planes - 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:​

Detailed Solution for Test: Intercept Form And Family Of Planes - Question 2

(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

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Test: Intercept Form And Family Of Planes - 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.

Test: Intercept Form And Family Of Planes - Question 4

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

Test: Intercept Form And Family Of Planes - Question 5

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

Detailed Solution for Test: Intercept Form And Family Of Planes - Question 5

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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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests