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  • Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 3 maths textbook solution

Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 3 maths textbook solution

Answers (1)

Answer:\frac{3}{2}\left ( \hat{i}+\hat{j} \right )

Given: The vector\overrightarrow{b}=3\hat{i}+4\hat{k}is to be written as the sum of a vector\overrightarrow{a}parallel to\overrightarrow{a}=\hat{i}+\hat{j}and\overrightarrow{\beta } 

perpendicular to \overrightarrow{a}. Then \overrightarrow{a}=

Hint: If\overrightarrow{a }\parallel \overrightarrow{a }\Rightarrow \overrightarrow{a }and if \overrightarrow{a }\perp \overrightarrow{a }\Rightarrow \overrightarrow{a }.\overrightarrow{a }=0

Explanation:

              Here \overrightarrow{a}=\hat{i}+\hat{j},  \overrightarrow{b}=3\hat{i}+4\hat{k}  

              Also \overrightarrow{b}=\overrightarrow{a}+\overrightarrow\beta                                                         ……………. (1)    where \overrightarrow{a}\parallel \overrightarrow{a}and \overrightarrow{\beta }\perp \overrightarrow{a}

As \vec{\alpha} \| \vec{a} \Rightarrow \vec{\alpha}=\lambda \vec{a} \Rightarrow \vec{\alpha}=\lambda(\hat{i}+\hat{j}) \Rightarrow \vec{\alpha}=\lambda \hat{i}+\lambda \hat{j} \quad[\because \vec{a} \| \vec{b} \Rightarrow \vec{a}=k \vec{b}]

By (1)

              \begin{aligned} &\vec{\beta}=\vec{b}-\vec{\alpha} \\ &\Rightarrow \vec{\beta}=3 \hat{i}+4 \hat{k}-\lambda \hat{i}-\lambda \hat{j} \\ &\Rightarrow \vec{\beta}=(3-\lambda) \hat{i}-\lambda \hat{j}+4 \hat{k} \end{aligned}

Now \vec{\beta}.\vec{a}=0                                                             ………….. (2)                     \because \vec{\beta}\perp \vec{a}

              \begin{aligned} &\Rightarrow \vec{\beta} \cdot \vec{a}=(3-\lambda)(1)-\lambda(1)+4(0) \\ &=3-\lambda-\lambda \\ &=3-2 \lambda \end{aligned}

                             

              \begin{aligned} &\because \text { By (2) } \quad 3-2 \lambda=0 \Rightarrow 2 \lambda=3 \Rightarrow \lambda=\frac{3}{2} \\ &\Rightarrow \vec{\alpha}=\frac{3}{2} \vec{a}=\frac{3}{2}(\hat{i}+\hat{j}) \\ &\Rightarrow \vec{\alpha}=\frac{3}{2}(\hat{i}+\hat{j}) \end{aligned}    

                             

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