#### Need solution for RD Sharma maths class 12 chapter 24 vector or cross product exercise multiple choice questions question 1 maths textbook solution

Answer:$2\overrightarrow{a}^{2}$

Given: If $\overrightarrow{a}$ is any vector, then $(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}$

Hint: You must know about the vector products of orthonormal triad of unit vectors.

Explanation: Let  $\overrightarrow{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$

Now,

\begin{aligned} &\vec{a} \times \hat{i}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{i} \\ &=a_{1}(\hat{i} \times \hat{i})+a_{2}(\hat{j} \times \hat{i})+a_{3}(\hat{k} \times \hat{i}) \quad[\therefore \hat{i} \times \hat{i}=0 ; \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{i}=\hat{j}] \end{aligned}

\begin{aligned} &=a_{1}(0)+a_{2}(-\hat{k})+a_{3}(\hat{j}) \\ &=0-a_{2} \hat{k}+a_{3} \hat{j} \\ &=a_{3} \hat{j}-a_{2} \hat{k} \end{aligned}

On squaring,

\begin{aligned} &\begin{aligned} &\left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=\left(a_{3} \hat{j}-a_{2} \hat{k}\right)\left(a_{3} \hat{j}-a_{2} \hat{k}\right) \\ &=\left(a_{3} \hat{j}\right)^{2}+\left(a_{2} \hat{k}\right)^{2}-2 a_{3} \hat{j} \cdot a_{2} \hat{k} \end{aligned} \quad\left[\begin{array}{l} \because \hat{j} \cdot \hat{j}=1 \\ \hat{k} \cdot \hat{k}=1 \\ \hat{j} \cdot \hat{k}=0 \end{array}\right]\\ &=a_{3}^{2}(1)+a_{2}^{2}(1)-2 a_{2} a_{3}(0)\\ &\left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=a_{2}^{2}+a_{3}^{2} \end{aligned}

$\begin{gathered} =a_{3}^{2}(1)+a_{2}^{2}(1)-2 a_{2} a_{3}(0) \\ \left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=a_{2}^{2}+a_{3}^{2} \end{gathered}$             …………… (1)

Similarly,

\begin{aligned} &\vec{a} \times \hat{j}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{j} \\ &=a_{1}(\hat{i} \times \hat{j})+a_{2}(\hat{j} \times \hat{j})+a_{3}(\hat{k} \times \hat{j}) \quad\left[\begin{array}{l} \because \hat{i} \times \hat{j}=\hat{k}, \\ \hat{j} \times \hat{j}=0, \\ \hat{k} \times \hat{j}=-\hat{i} \end{array}\right] \\ &=a_{1}(\hat{k})+a_{2}(0)+a_{3}(-\hat{i}) \\ &=a_{1} \hat{k}-a_{3} \hat{i} \end{aligned}        ………………… (2)

Again

\begin{aligned} &\vec{a} \times \hat{k}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{k} \\ &=a_{1}(\hat{i} \times \hat{k})+a_{2}(\hat{j} \times \hat{k})+a_{3}(\hat{k} \times \hat{k}) \quad\left[\begin{array}{l} \hat{i} \times \hat{k}=-\hat{j} \\ \hat{j} \times \hat{k}=\hat{i} \\ \hat{k} \times \hat{k}=0 \end{array}\right] \\ &=a_{1}(-\hat{j})+a_{2}(\hat{i})+a_{3}(0) \end{aligned}

\begin{aligned} &=a_{1}\hat{j}+a_{2}\hat{i} \end{aligned}

On squaring

\begin{aligned} &\left(a_{2} \hat{i}-a_{1} \hat{j}\right)^{2}=\left(a_{2} \hat{i}-a_{1} \hat{j}\right)\left(a_{2} \hat{i}-a_{1} \hat{j}\right)\\ &=\left(a_{2} \hat{i}\right)^{2}+\left(a_{1} \hat{j}\right)^{2}-2 a_{1} a_{2}(\hat{i} . \hat{j})\left[\begin{array}{l} \because \hat{i} . \hat{i}=1 \\ \hat{j} \cdot \hat{j}=1 \\ \hat{i} \cdot \hat{j}=0 \end{array}\right]\\ &=a_{2}^{2}(1)+a_{1}^{2}(1)-2 a_{1} a_{2}(0)\\ &=a_{2}^{2}+a_{1}^{2} \end{aligned}      ……………… (3)

Now,

\begin{aligned} (\vec{a} \times \hat{i})^{2}+&(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2} \\ &=a_{2}^{2}+a_{3}^{2}+a_{1}^{2}+a_{3}^{2}+a_{2}^{2}+a_{1}^{2}[\because \operatorname{from}(1),(2),(3)] \\ &=2\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)=2 \vec{a}^{2} \end{aligned}

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Answer:$2\overrightarrow{a}^{2}$

Given: If $\overrightarrow{a}$ is any vector, then $(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}$

Hint: You must know about the vector products of orthonormal triad of unit vectors.

Explanation: Let  $\overrightarrow{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$

Now,

\begin{aligned} &\vec{a} \times \hat{i}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{i} \\ &=a_{1}(\hat{i} \times \hat{i})+a_{2}(\hat{j} \times \hat{i})+a_{3}(\hat{k} \times \hat{i}) \quad[\therefore \hat{i} \times \hat{i}=0 ; \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{i}=\hat{j}] \end{aligned}

\begin{aligned} &=a_{1}(0)+a_{2}(-\hat{k})+a_{3}(\hat{j}) \\ &=0-a_{2} \hat{k}+a_{3} \hat{j} \\ &=a_{3} \hat{j}-a_{2} \hat{k} \end{aligned}

On squaring,

\begin{aligned} &\begin{aligned} &\left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=\left(a_{3} \hat{j}-a_{2} \hat{k}\right)\left(a_{3} \hat{j}-a_{2} \hat{k}\right) \\ &=\left(a_{3} \hat{j}\right)^{2}+\left(a_{2} \hat{k}\right)^{2}-2 a_{3} \hat{j} \cdot a_{2} \hat{k} \end{aligned} \quad\left[\begin{array}{l} \because \hat{j} \cdot \hat{j}=1 \\ \hat{k} \cdot \hat{k}=1 \\ \hat{j} \cdot \hat{k}=0 \end{array}\right]\\ &=a_{3}^{2}(1)+a_{2}^{2}(1)-2 a_{2} a_{3}(0)\\ &\left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=a_{2}^{2}+a_{3}^{2} \end{aligned}

$\begin{gathered} =a_{3}^{2}(1)+a_{2}^{2}(1)-2 a_{2} a_{3}(0) \\ \left(a_{3} \hat{j}-a_{2} \hat{k}\right)^{2}=a_{2}^{2}+a_{3}^{2} \end{gathered}$             …………… (1)

Similarly,

\begin{aligned} &\vec{a} \times \hat{j}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{j} \\ &=a_{1}(\hat{i} \times \hat{j})+a_{2}(\hat{j} \times \hat{j})+a_{3}(\hat{k} \times \hat{j}) \quad\left[\begin{array}{l} \because \hat{i} \times \hat{j}=\hat{k}, \\ \hat{j} \times \hat{j}=0, \\ \hat{k} \times \hat{j}=-\hat{i} \end{array}\right] \\ &=a_{1}(\hat{k})+a_{2}(0)+a_{3}(-\hat{i}) \\ &=a_{1} \hat{k}-a_{3} \hat{i} \end{aligned}        ………………… (2)

Again

\begin{aligned} &\vec{a} \times \hat{k}=\left(a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\right) \times \hat{k} \\ &=a_{1}(\hat{i} \times \hat{k})+a_{2}(\hat{j} \times \hat{k})+a_{3}(\hat{k} \times \hat{k}) \quad\left[\begin{array}{l} \hat{i} \times \hat{k}=-\hat{j} \\ \hat{j} \times \hat{k}=\hat{i} \\ \hat{k} \times \hat{k}=0 \end{array}\right] \\ &=a_{1}(-\hat{j})+a_{2}(\hat{i})+a_{3}(0) \end{aligned}

\begin{aligned} &=a_{1}\hat{j}+a_{2}\hat{i} \end{aligned}

On squaring

\begin{aligned} &\left(a_{2} \hat{i}-a_{1} \hat{j}\right)^{2}=\left(a_{2} \hat{i}-a_{1} \hat{j}\right)\left(a_{2} \hat{i}-a_{1} \hat{j}\right)\\ &=\left(a_{2} \hat{i}\right)^{2}+\left(a_{1} \hat{j}\right)^{2}-2 a_{1} a_{2}(\hat{i} . \hat{j})\left[\begin{array}{l} \because \hat{i} . \hat{i}=1 \\ \hat{j} \cdot \hat{j}=1 \\ \hat{i} \cdot \hat{j}=0 \end{array}\right]\\ &=a_{2}^{2}(1)+a_{1}^{2}(1)-2 a_{1} a_{2}(0)\\ &=a_{2}^{2}+a_{1}^{2} \end{aligned}      ……………… (3)

Now,

\begin{aligned} (\vec{a} \times \hat{i})^{2}+&(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2} \\ &=a_{2}^{2}+a_{3}^{2}+a_{1}^{2}+a_{3}^{2}+a_{2}^{2}+a_{1}^{2}[\because \operatorname{from}(1),(2),(3)] \\ &=2\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)=2 \vec{a}^{2} \end{aligned}