MHT CET · Maths · Three Dimensional Geometry
The plane through the intersection of planes \(x+y+z=1\) and \(2 x+3 y-z+4=0\) and parallel to Y-axis also passes through the point
- A \((3,3,-1)\)
- B \((-3,0,1)\)
- C \((3,2,1)\)
- D \((-3,0,-1)\)
Answer & Solution
Correct Answer
(C) \((3,2,1)\)
Step-by-step Solution
Detailed explanation
Equation of plane through the intersection of given planes is
\(\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \quad \ldots \text { (i) } \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}\)
Since the plane is parallel to \(\mathrm{Y}\)-axis.
\(\begin{aligned}
\therefore \quad & 1+3 \lambda=0 \\
& \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}\)
Substituting \(\lambda=\frac{-1}{3}\) in (i), we get
\(\begin{aligned}
& (x+y+z-1)-\frac{1}{3}(2 x+3 y-z+4)=0 \\
& \Rightarrow x+4 z-7=0
\end{aligned}\)
Point \((3,2,1)\) satisfies this equation.
\(\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \quad \ldots \text { (i) } \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}\)
Since the plane is parallel to \(\mathrm{Y}\)-axis.
\(\begin{aligned}
\therefore \quad & 1+3 \lambda=0 \\
& \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}\)
Substituting \(\lambda=\frac{-1}{3}\) in (i), we get
\(\begin{aligned}
& (x+y+z-1)-\frac{1}{3}(2 x+3 y-z+4)=0 \\
& \Rightarrow x+4 z-7=0
\end{aligned}\)
Point \((3,2,1)\) satisfies this equation.
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- Let \(f: R \rightarrow R\) and \(g: R \rightarrow R\) be continuous functions. Then the value of the integral
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \mathrm{d} x\) isMHT CET 2023 Medium - If the points \(A(3,2,1), B(4, x, 5), C(4,2,-2)\) and \(D(6,5,-1)\) are coplanar, then \(x\) has the valueMHT CET 2022 Easy
- Let \(a\) and \(b\) two non-zero real numbers. The equation
\(\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)\)
Four straight lines, when \(c=0\) and \(a, b\) are of the same sign.MHT CET 2022 Easy - If with thenMHT CET 2018 Medium
- If \(\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \overline{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\), then a vector \(\bar{d}\) which is parallel to vector \(\bar{a} \times \bar{b}\) and which \(\overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=15\), isMHT CET 2023 Medium
- If the curve \(y=a x^2-6 x+b\) passes through \((0,4)\) and has its tangent parallel to the x -axis at \(x=\frac{3}{2}\), then the value of \(a\) and b respectively areMHT CET 2025 Medium
More PYQs from MHT CET
- Willemite isMHT CET 2009 Hard
- If the sum of mean and variance of a binomial distribution for 5 trials is 1.8 , then probability of a success isMHT CET 2021 Easy
- If and are centroid, orthocenter and circumcenter of a triangle and thenMHT CET 2016 Medium
- Identify product ' \(B\) ' in the following reaction.
Sodium phenoxide \(\xrightarrow[6 \text { atm }]{\mathrm{CO}_2, 398 \mathrm{~K}} \mathrm{~A} \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}} \mathrm{B}\)MHT CET 2024 Hard - One of the human embryonic developmental stages has been shown in the diagram. What do the A and B indicate?
MHT CET 2020 Medium - In the following circuit, the reading in the ammeter is
MHT CET 2024 Easy