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  • Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 5 Maths.

Explain Solution RD Sharma Class 12 Chapter Algebra of Vectors Exercise 22.6 Question 5 Maths.

Answers (1)

Answer:
|3 \vec{a}-2 \vec{b}+4 \vec{c}| \text { will be } \sqrt{398}
Hint:

Use vector magnitude formula  |\vec{a}|=\sqrt{a^{2}+b^{2}+c^{2}}

Given:

\vec{a}=3 \hat{\imath}-\hat{\jmath}-4 \hat{k}, \vec{b}=-2 \hat{\imath}+4 \hat{\jmath}-3 \hat{k} \text { and } \overrightarrow{\mathrm{c}}=\hat{\imath}+2 \hat{\jmath}-\hat{k}

Solution:

We want to find the magnitude of vector  3 \vec{a}-2 \vec{b}+4 \vec{c}

So,

\begin{aligned} &3 \vec{a}-2 \vec{b}+4 \vec{c}=3(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})-2(-2 \hat{\imath}+4 \hat{\jmath}-3 \hat{k})+4(\hat{\imath}+2 \hat{\jmath}-\hat{k}) \\ &3 \vec{a}-2 \vec{b}+4 \vec{c}=9 \hat{i}-3 \hat{j}-12 \hat{k}+4 \hat{\imath}-8 \hat{\jmath}+6 \hat{k}+4 \hat{\imath}+8 \hat{j}-4 \hat{k} \\ &3 \vec{a}-2 \vec{b}+4 \vec{c}=17 \hat{\imath}-3 \hat{\jmath}-10 \hat{k} \end{aligned}

If a vector is given by:

\vec{A}=a \hat{i}-b \hat{j}-c \hat{k}  then the magnitude of vector is generally by |\vec{a}|   which is equal to \sqrt{a^{2}+b^{2}+c^{2}}

\begin{array}{r} |3 \vec{a}-2 \vec{b}+4 \vec{c}|=\sqrt{17^{2}+(-3)^{2}+(-10)^{2}} \\\\ =\sqrt{289+9+100}=\sqrt{398} \end{array}

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