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  • Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 4 Maths.

Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 4 Maths.

Answers (1)

Answer:

The unit vector parallel to diagonals will be  \frac{1}{\sqrt{6}}(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \& \frac{-\hat{\imath}+\hat{k}}{\sqrt{2}}

Hint:

Use vector magnitude formula.

Given:

\hat{\imath}+\hat{\jmath}-\hat{k} \text { and }-2 \hat{\imath}+\hat{\jmath}+2 \hat{k}

Solution:          

    

Side BC parallel to:  A \vec{D}=\vec{b}, A \vec{B}=\vec{a}

So, resultant vector   A \vec{C}=\vec{c}=\vec{a}+\vec{b}

So, vector

\begin{aligned} &\vec{c}=\vec{a}+\vec{b}=(\hat{\imath}+\hat{\jmath}-\hat{k})+(-2 \hat{\imath}+\hat{\jmath}+2 \hat{k}) \\\\ &\vec{c}=\hat{\imath}+\hat{\jmath}-\hat{k}-2 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\\\ &\vec{c}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \end{aligned}

So, unit vector along the diagonal of parallelogram is

\begin{aligned} &\vec{c}=\frac{\vec{c}}{|\vec{c}|}=\frac{-\hat{\imath}+2 \hat{\jmath}+\hat{k}}{\sqrt{(-1)^{2}+2^{2}+1^{2}}} \\\\ &A \vec{C}=\vec{c}=\frac{-\hat{\imath}+2 \hat{\jmath}+\hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}}(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \end{aligned}

Similarly   \begin{aligned} & B \vec{D}=\vec{b}-\vec{a}=-3 \hat{\imath}+3 \hat{k} \\\\ &\end{aligned}

\widehat{B D}=\frac{-3 \hat{\imath}+3 \hat{k}}{\sqrt{-3^{2}+3^{2}}}=\frac{-\hat{\imath}+\hat{k}}{\sqrt{2}}

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