Question

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 parallel to y-axis also passes through the point:
1. (-3, 0, -1)
2. (-3, 1, 1)
3. (3, 3, -1)
4. (3, 2, 1)

Answer 1

Correct Answer - Option 4 : (3, 2, 1)

Since, equation of plane through intersection of planes

x + y + z = 1       ----(1)      

2x + 3y – z + 4 = 0        ----(2)

Equation of the plane is:

(A₁ x + B₁ y + C₁ z – d₁) + λ(A₂ x + B₂ y + C₂ z – d₂) = 0       ----(3)

Using equation (1) and (2) in the equation (3)

⇒ (x + y + z – 1) + λ(2x + 3y – z + 4) = 0

⇒ (x + y + z – 1) + (2λx + 3λy – λz + 4λ) = 0

⇒ (x + 2λx) + (y + 3λy) + (z – λz) + (-1 + 4λ) = 0

(1 + 2λ)x + (1 + 3λ)y + (1 – λ)z + (-1 + 4λ) = 0        ----(4)

But the above plane is parallel to y-axis then its normal plane is perpendicular to y-axis.

\(\Rightarrow \left( {\left( {1 + 2{\rm{\lambda }}} \right){\rm{\hat i}} + \left( {1 + 3{\rm{\lambda }}} \right){\rm{\hat j}} + \left( {1 - {\rm{\lambda }}} \right){\rm{\hat k}}} \right).\left( {0{\rm{\hat i}} + {\rm{\hat j}} + 0{\rm{\hat k}}} \right) = 0\)

⇒ [(1 + 2λ) × 0] + [(1 + 3λ) × 1] + [(1 – λ) × 0] = 0

⇒ 1 + 3λ = 0

⇒ 1 = -3λ

\(\therefore {\rm{\lambda }} = - \frac{1}{3}\)

By substituting λ = -3 in equation (4),

\(\Rightarrow \left( {1 + 2\left( { - \frac{1}{3}} \right)} \right){\rm{x}} + \left( {1 + 3\left( { - \frac{1}{3}} \right)} \right){\rm{y}} + \left( {1 - \left( { - \frac{1}{3}} \right)} \right){\rm{z}} + \left( { - 1 + 4\left( { - \frac{1}{3}} \right)} \right) = 0\)

\(\Rightarrow \left( {1 - \frac{2}{3}} \right){\rm{x}} + \left( {1 - 1} \right){\rm{y}} + \left( {1 + \frac{1}{3}} \right){\rm{z}} + \left( { - 1 - \frac{4}{3}} \right) = 0\)

\(\Rightarrow \left( {\frac{{3 - 2}}{3}} \right){\rm{x}} + 0{\rm{y}} + \left( {\frac{{3 + 1}}{3}} \right){\rm{z}} + \left( {\frac{{ - 3 - 4}}{3}} \right) = 0\)

\(\Rightarrow \left( {\frac{1}{3}} \right){\rm{x}} + \left( {\frac{4}{3}} \right){\rm{z}} - \left( {\frac{7}{3}} \right) = 0\)

Hence, the equation of required plane is

∴ x + 4z – 7 = 0

Now, we need to substitute the points given in the option to find which point is satisfying the above equation. The point which satisfies the equation is the point through which the plane is passing.

Option (a):

(-3, 0, -1) ⇒ -3 + 4(-1) – 7 ≠ 0

Option (b):

(-3, 1, 1) ⇒ -3 + 4(1) – 7 ≠ 0

Option (c):

(3, 3, -1) ⇒ 3 + 4(-1) – 7 ≠ 0

Option (d):

(3, 2, 1) ⇒ 3 + 4(1) – 7 = 0

Therefore, the plane passes through the point (3, 2, 1).

Thus, the correct answer is option (d).