CUET · MATHS · PYQ PAPER 2023

Let \(P\) be the point of intersection of the line \(\frac{x+5}{6}=\frac{y+4}{1}=\frac{3-z}{1}\) and the plane \(x+y+2 z=2\).
If \(Q\) is the point \((-3,3,4)\), then \((P Q)^2\) is equal to

A 66
B 60
C 56
D 52

Verified Solution

Answer & Solution

Correct Answer

(C) 56

Step-by-step Solution

Detailed explanation

Line: \(x=6\lambda-5\), \(y=\lambda-4\), \(z=3-\lambda\) Substitute into plane: \((6\lambda-5)+(\lambda-4)+2(3-\lambda)=2\) \(5\lambda-3=2 \Rightarrow 5\lambda=5 \Rightarrow \lambda=1\) Point \(P\): \((6(1)-5, 1-4, 3-1) = (1, -3, 2)\) Point \(Q\): \((-3,3,4)\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List-I with List-II

List-I	List-II
(A) The vectors \(\lambda \hat{i}+\hat{j}+2 \hat{k}\) and \(\hat{i}+\lambda \hat{j}+\hat{k}\) are perpendicular if \(\lambda\) is equal to	(I) 1
(B) The vectors \(3 \hat{i}+6 \hat{j}-\hat{k}\) and \(2 \hat{i}+\lambda \hat{j}-2 \hat{k}\) are collinear if \(\lambda\) is equal to	(II) -1
(C) The number of vectors of unit-length which are perpendicular to both the vectors \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=3 \hat{i}-\hat{j}+5 \hat{k}\) is	(III) \(2 / 3\)
(D) If \(\|\vec{a}\|=1\) and \(\vec{a}+\vec{b}=\overrightarrow{0}\), then \(\|\vec{b}\|\) is equal to	(IV) 2

Choose the correct answer from the options given below :

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Let \(P\) be the point of intersection of the line \(\frac{x+5}{6}=\frac{y+4}{1}=\frac{3-z}{1}\) and the plane \(x+y+2 z=2\).If \(Q\) is the point \((-3,3,4)\), then \((P Q)^2\) is equal to

Step-by-step Solution

Let \(P\) be the point of intersection of the line \(\frac{x+5}{6}=\frac{y+4}{1}=\frac{3-z}{1}\) and the plane \(x+y+2 z=2\).
If \(Q\) is the point \((-3,3,4)\), then \((P Q)^2\) is equal to