#### Need solution for RD sharma maths class 12 chapter 24 Vector or cross product exercise Fill in the blanks question 7

$\lambda = 2$

Hint: Use formula

$\left | \vec{a} \times \vec{i }\right |^{2}=\left | \vec{a} \times \vec{i }\right |.\left | \vec{a} \times \vec{i }\right |$

Given:

$\left | \vec{r} \times \vec{i }\right |^{2}+\left | \vec{r} \times \vec{j}\right |^{2}+\left | \vec{r} \times \vec{k}\right |^{2}=\lambda |r|^{2}$

Solving

$\left | \vec{r} \times \vec{i }\right |^{2}=\left (\vec{r} \times \vec{i}\right ).\left ( \vec{r} \times \vec{i}\right )$

$=\begin{vmatrix} \hat{r}.\hat{r} & \hat{r}.\hat{i} \\ \hat{i}.\hat{r} & \hat{i}.\hat{i} \end{vmatrix}\\ =\begin{vmatrix} \hat{r}.\hat{r} & \hat{r}.\hat{i} \\ \hat{i}.\hat{r} & 1 \end{vmatrix}$$=\begin{vmatrix} \hat{r}.\hat{r} & \hat{r}.\hat{i} \\ \hat{i}.\hat{r} & \hat{i}.\hat{i} \end{vmatrix}\\ =\begin{vmatrix} \hat{r}.\hat{r} & \hat{r}.\hat{i} \\ \hat{i}.\hat{r} & 1 \end{vmatrix}$

$=|\vec{r}|^{2}-(\vec{r}.\vec{i})^{2}$............................(1)

Similarly

$|\vec{r} \times \vec{j}|^{2}=|\vec{r}|^{2}-(\vec{r} \cdot \vec{j})^{2}$ ................(2)

$|\vec{r} \times \vec{k}|^{2}=|\vec{r}|^{2}-(\vec{r} \cdot \vec{k})^{2}$.................(3)

we get,

\begin{aligned} &|\vec{r} \times \vec{i}|^{2}+|\vec{r} \times \vec{j}|^{2}+|\vec{r} \times \vec{k}|^{2} \\ &=\left[|\vec{r}|^{2}-|\vec{r} \cdot \vec{i}|^{2}\right]+\left[|\vec{r}|^{2}-|\vec{r} \cdot \vec{j}|^{2}\right]+\left[|\vec{r}|^{2}-|\vec{r} \cdot \vec{k}|^{2}\right] \\ \end{aligned}

\begin{aligned} &\vec{r} .\hat{i}=(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}+0 \hat{j}+0 \hat{k}) \\ &=(\hat{i})^{2} \\ \end{aligned}

\begin{aligned} &\vec{r} \cdot \vec{j}=(\hat{i}+\hat{j}+\hat{k}) \cdot(0 \hat{i}+\hat{j}+0 \hat{k}) \\ &=(\hat{j})^{2} \\ \end{aligned}

\begin{aligned} &\vec{r} \cdot \vec{k}=(\hat{i}+\hat{j}+\hat{k}) \cdot(0 \hat{i}+0 \hat{j}+\hat{k}) \\ &=(\hat{k})^{2} \\ \end{aligned}

\begin{aligned} &\Rightarrow\left[|\vec{r}|^{2}+|\vec{r}|^{2}+|\vec{r}|^{2}\right]-\left[|\vec{r} \cdot \vec{i}|^{2}+|\vec{r} \cdot \vec{j}|^{2}+|\vec{r} \vec{k}|^{2}\right] \\ \end{aligned}

\begin{aligned} &=3|\vec{r}|^{2}-\left[\left|\hat{i}^{2}+\hat{j}^{2}+\hat{k}^{2}\right|\right] \\ &=3|\vec{r}|^{2}-|\vec{r}|^{2} \\ &=2|\vec{r}|^{2} \\ \end{aligned}

We have been given,

\begin{aligned}&|\vec{r} \times \vec{i}|^{2}+|\vec{r} \times \vec{j}|^{2}+|\vec{r} \times \vec{k}|^{2}=\lambda|\vec{r}|^{2} \\ &\lambda=2 \end{aligned}