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MHT CET 202621 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
21 April
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

MHT CET 2026 · 21 April · Evening Shift · Physics · Mathematics in Physics

Out of the following vectors, identify the one which is normal to...

Out of the following vectors, identify the one which is normal to the force, F⃗=4i^−3j^\vec{F} = 4\hat{i} - 3\hat{j}.

Options

  1. A−6i^-6\hat{i}
  2. B7i^7\hat{i}
  3. C3i^+4j^3\hat{i} + 4\hat{j}
  4. D4i^+3j^4\hat{i} + 3\hat{j}

Correct answer

C. 3i^+4j^3\hat{i} + 4\hat{j}

Step-by-step solution

Two vectors are normal to each other if their dot product is zero.

Given force vector F⃗=4i^−3j^\vec{F} = 4\hat{i} - 3\hat{j}.

Checking the dot product of F⃗\vec{F} with each of the given vectors:

For option (1): (4i^−3j^)⋅(−6i^)=(4)(−6)+(−3)(0)=−24≠0(4\hat{i} - 3\hat{j}) \cdot (-6\hat{i}) = (4)(-6) + (-3)(0) = -24 \neq 0

For option (2): (4i^−3j^)⋅(7i^)=(4)(7)+(−3)(0)=28≠0(4\hat{i} - 3\hat{j}) \cdot (7\hat{i}) = (4)(7) + (-3)(0) = 28 \neq 0

For option (3): (4i^−3j^)⋅(3i^+4j^)=(4)(3)+(−3)(4)=12−12=0(4\hat{i} - 3\hat{j}) \cdot (3\hat{i} + 4\hat{j}) = (4)(3) + (-3)(4) = 12 - 12 = 0

For option (4): (4i^−3j^)⋅(4i^+3j^)=(4)(4)+(−3)(3)=16−9=7≠0(4\hat{i} - 3\hat{j}) \cdot (4\hat{i} + 3\hat{j}) = (4)(4) + (-3)(3) = 16 - 9 = 7 \neq 0

Since the dot product is zero for the vector 3i^+4j^3\hat{i} + 4\hat{j}, it is normal to the given force vector.
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