Question:
If \( \vec{A} = a_1 \hat{i} + a_2 \hat{j} \) and \( \vec{B} = b_1 \hat{i} + b_2 \hat{j} \) are perpendicular to each other, then
If \( \vec{A} = a_1 \hat{i} + a_2 \hat{j} \) and \( \vec{B} = b_1 \hat{i} + b_2 \hat{j} \) are perpendicular to each other, then
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For perpendicular vectors, their dot product is zero. Use this condition to derive relations between the components.
Updated On: Jan 26, 2026
- \( \frac{b_2}{a_1} = - \frac{a_2}{b_1} \)
- \( \frac{a_1}{b_2} = \frac{a_2}{b_1} \)
- \( \frac{b_2}{a_1} = \frac{a_2}{b_1} \)
- \( \frac{a_1}{b_2} = - \frac{a_2}{b_1} \)
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The Correct Option is D
Solution and Explanation
Step 1: Using the perpendicularity condition.
When two vectors are perpendicular, their dot product is zero. The dot product of \( \vec{A} \) and \( \vec{B} \) is given by: \[ \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 = 0 \] This implies: \[ a_1 b_1 = -a_2 b_2 \] Step 2: Solving for the ratio.
Rearranging the equation, we get: \[ \frac{a_1}{b_2} = -\frac{a_2}{b_1} \] Step 3: Conclusion.
Thus, the correct answer is (D), \( \frac{a_1}{b_2} = - \frac{a_2}{b_1} \).
When two vectors are perpendicular, their dot product is zero. The dot product of \( \vec{A} \) and \( \vec{B} \) is given by: \[ \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 = 0 \] This implies: \[ a_1 b_1 = -a_2 b_2 \] Step 2: Solving for the ratio.
Rearranging the equation, we get: \[ \frac{a_1}{b_2} = -\frac{a_2}{b_1} \] Step 3: Conclusion.
Thus, the correct answer is (D), \( \frac{a_1}{b_2} = - \frac{a_2}{b_1} \).
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