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Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 18 Maths.

Answers (1)

Answer:

\frac{15}{\sqrt{10}} \hat{\imath}+\frac{5}{\sqrt{10}} \hat{\jmath}

Hint:

A vector having magnitude 1 and parallel to  \vec{a}=1 \frac{\vec{a}}{|\vec{a}|}

Given:

\vec{a}=(2 \hat{\imath}+3 \hat{\jmath}-\hat{k}), \vec{b}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})

Solution:

If  \vec{c}  is the resultant of  \vec{a} \text { and } \vec{b}

\begin{aligned} &\vec{c}=(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})+(\hat{\imath}-2 \hat{\jmath}+\hat{k}) \\\\ &\vec{c}=3 \hat{\imath}+\hat{\jmath}+0 \hat{k} \end{aligned}

Now, a vector having magnitude 5 and parallel to \vec{c}

\begin{aligned} &\frac{5 \vec{c}}{|\vec{c}|}=\frac{5(3 \hat{\imath}+\hat{j})}{\sqrt{3^{2}+1^{2}}} \\\\ &=\frac{15}{\sqrt{10}} \hat{\imath}+\frac{5}{\sqrt{10}} \hat{\jmath} \end{aligned}

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