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  • Provide solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 6

Provide solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 6

Answers (1)

0

Hint:

  Find \alpha

Given:

If the angle between vector such that \hat{i}+\hat{k} and \hat{i}- \hat{j}+\alpha \hat{k}  is \frac{\pi}{3}.

Find \alpha

Solution:

\begin{aligned} &\cos \theta=\frac{\vec{a} \vec{b}}{|\vec{a}| \cdot|\vec{b}|^{\prime}} \theta=\frac{\pi}{3}\\ \end{aligned}

\begin{aligned} &\cos \theta=\frac{(\hat{\imath}+\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+\alpha \hat{k})}{\sqrt{1^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+\alpha^{2}}}\\ \end{aligned}

\begin{aligned} &\cos \frac{\pi}{3}=\frac{(\hat{\imath}+\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+\alpha \hat{k})}{\sqrt{2} \cdot \sqrt{2+\alpha^{2}}}\\ \end{aligned}

\begin{aligned} &\frac{1}{2}=\frac{(1+\alpha)}{\sqrt{2} \cdot \sqrt{2+\alpha^{2}}}\\ \end{aligned}

\begin{aligned} &\frac{1}{4}=\frac{(1+\alpha)^{2}}{2\left(2+\alpha^{2}\right)} \\ \end{aligned}..............…squaring the equation

\begin{aligned} &2(1+\alpha)^{2}=2+\alpha^{2}\\ &2\left(1+\alpha^{2}+2 \alpha\right)=2+\alpha^{2}\\ &2+2 \alpha^{2}+4 \alpha=2+\alpha^{2}\\ \end{aligned}

\begin{aligned} &\alpha^{2}+4 \alpha=0\\ &\alpha(\alpha+4)=0\\ &\alpha=0, \alpha=-4\\ \end{aligned}

Hence \begin{aligned} & \alpha=0\end{aligned}
 

      Hence α=0

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