SATHEE: Vectors 1 Question 7

Vectors 1 Question 7

8. Two adjacent sides of a parallelogram $A B C D$ are given by $\vec{A B} = 2 \hat{i} + 10 \hat{j} + 11 \hat{k}$ and $\vec{A D} = - \hat{i} + 2 \hat{j} + 2 \hat{k}$ . The side $A D$ is rotated by an acute angle $α$ in the plane of the parallelogram so that $A D$ becomes $A D$ ‘. If $A D$ ’ makes a right angle with the side $A B$ , then the cosine of the angle $α$ is given by

(2010)

(a) $\frac{8}{9}$

(b) $\frac{\sqrt{17}}{9}$

(d) $\frac{4 \sqrt{5}}{9}$

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Answer:

Correct Answer: 8. (b)

Solution:

$\vec{A B} = 2 \hat{i} + 10 \hat{j} + 11 \hat{k}, \vec{A D} = - \hat{i} + 2 \hat{j} + 2 \hat{k}$

Angle ’ $θ$ ’ between $\vec{A B}$ and $\vec{A D}$ is

$\begin{aligned} \cos (θ) = | \frac{\vec{A B} \cdot \vec{A D}}{| \vec{A B} | | \vec{A D} |} | = | \frac{- 2 + 20 + 22}{(15) (3)} | = \frac{8}{9} \\ \Rightarrow \sin (θ) = \frac{\sqrt{17}}{9} \end{aligned}$

Since, $α + θ = 90^{\circ}$

$∴ \cos (α) = \cos (90^{\circ} - θ) = \sin (θ) = \frac{\sqrt{17}}{9}$

Vectors 1 Question 7

Answer:

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Vectors 1 Question 7

Answer:

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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