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  • Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 11 maths textbook solution

Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 11 maths textbook solution

Answers (1)

Answer: \frac{2}{\sqrt{7}}

Given: \theta is the angle between the vectors 2\hat{i}-2\hat{j}+4\hat{k} & 3\hat{i}+\hat{j}+2\hat{k}, then \sin \theta

Hint: Using \sin \theta =\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left | \overrightarrow{a}\times \overrightarrow{b} \right |}

Solution:

              \vec{a}=2 \hat{i}-2 \hat{j}+4 \hat{k} ; \vec{b}=3 \hat{i}+\hat{j}+2 \hat{k} \\ \Rightarrow|\vec{a}|=\sqrt{2^{2}+(-2)^{2}+(4)^{2}}=\sqrt{4+4+16}=\sqrt{24} \\ \Rightarrow|\vec{b}|=\sqrt{(3)^{2}+(1)^{2}+(2)^{2}}=\sqrt{9+1+4}=\sqrt{14} \

             \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 4 \\ 3 & 1 & 2 \end{array}\right| \\

                           =\hat{i}(-4-4)-\hat{j}(4-12)+\hat{k}(2+6) \\

                          =-8 \hat{i}+8 \hat{j}+8 \hat{k} \\

|\vec{a} \times \vec{b}|=\sqrt{(-8)^{2}+(8)^{2}+(8)^{2}}=\sqrt{64+64+64}=\sqrt{192} \\ \qquad

\because \sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{\sqrt{192}}{\sqrt{24} \sqrt{14}}=\frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3}}{\sqrt{2 \times 2 \times 2 \times 3} \sqrt{2 \times 7}}=\frac{\sqrt{4}}{\sqrt{7}}=\frac{2}{\sqrt{7}}

                      \qquad \sin \theta=\frac{2}{\sqrt{7}}

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