SATHEE: Kinematics Question 441

Kinematics Question 441

Question: Unit vector perpendicular to vector $\vec{A} = 3 \hat{i} + \hat{j}$ and $\vec{B} = 2 \hat{i} - \hat{j} - 5 \hat{k}$ both it’s

Options:

A) $\pm \frac{3 j - 2 \hat{k}}{\sqrt{11}}$

B) $\pm \frac{(\hat{i} - 3 \hat{j} + \hat{k})}{\sqrt{11}}$

C) $\pm \frac{- \hat{j} + 2 \hat{k}}{\sqrt{13}}$

D) $\pm \frac{\hat{i} + 3 \hat{j} - \hat{k}}{\sqrt{13}}$

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Answer:

Correct Answer: B

Solution:

[b] The unit vector in normal direction is $\hat{n} \pm \frac{\vec{A} \times \vec{B}}{| \vec{A} | | \vec{B} | \sin θ}$

Here, $\vec{A} \times \vec{B} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 2 & - 1 & - 5 \end{matrix} |$

$= - 5 \hat{i} + 15 \hat{j} - 5 \hat{k}$

$| \vec{A} | = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$

$| \vec{B} | = \sqrt{{(2)}^{2} + {(- 1)}^{2} + {(- 5)}^{2}} = \sqrt{30}$

$\cos θ = \frac{\vec{A} . \vec{B}}{| \vec{A} | | \vec{B} |} = \frac{1}{2 \sqrt{3}}$

$∴$ $\sin θ = \sqrt{1 - \cos^{2} θ} = \frac{\sqrt{11}}{2 \sqrt{3}}$

$∴$ $\hat{n} \pm \frac{- 5 \hat{i} + 15 \hat{j} - 5 \hat{k}}{\sqrt{10} . \sqrt{30} . \frac{\sqrt{11}}{2 \sqrt{3}}}$ $= \pm \frac{(\hat{i} - 3 \hat{j} + \hat{k})}{\sqrt{11}}$

Kinematics Question 441

Question: Unit vector perpendicular to vector $\vec{A} = 3 \hat{i} + \hat{j}$ and $\vec{B} = 2 \hat{i} - \hat{j} - 5 \hat{k}$ both it’s

Options:

Answer:

Solution:

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Kinematics Question 441

Question: Unit vector perpendicular to vector A→=3i^+j^ and B→=2i^−j^−5k^ both it’s

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: Unit vector perpendicular to vector $\vec{A} = 3 \hat{i} + \hat{j}$ and $\vec{B} = 2 \hat{i} - \hat{j} - 5 \hat{k}$ both it’s