Sankalp NEET Full Test-5 Question-17 Solution

Question: 17: If \vec{F}=2 \hat{i}+\hat{j}-\hat{k} and \vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k}, then the scalar and vector products of \vec{F} and \vec{r} have the magnitudes respectively as :

(1) 5, \sqrt{3}

(2) 4, \sqrt{5}

(3) 10, \sqrt{2}

(4) 10,2

Answer: Option (3)

Explanation:

First, we find the scalar (dot) product of the vectors.

The scalar product is given by

\vec{F}\cdot\vec{r} = F_x r_x + F_y r_y + F_z r_z

Substituting the given values,

\vec{F}\cdot\vec{r} = (2)(3) + (1)(2) + (-1)(-2) \vec{F}\cdot\vec{r} = 6 + 2 + 2 = 10

Hence, the magnitude of the scalar product is 10.

Now, we find the vector (cross) product of the vectors.

\vec{F}\times\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 3 & 2 & -2 \end{vmatrix}

Evaluating the determinant,

\vec{F}\times\vec{r} = \hat{i}[(1)(-2) - (-1)(2)] - \hat{j}[(2)(-2) - (-1)(3)] + \hat{k}[(2)(2) - (1)(3)] \vec{F}\times\vec{r} = \hat{i}( -2 + 2 ) - \hat{j}( -4 + 3 ) + \hat{k}( 4 - 3 ) \vec{F}\times\vec{r} = \hat{j} + \hat{k}

The magnitude of the vector product is

|\vec{F}\times\vec{r}| = \sqrt{1^2 + 1^2} = \sqrt{2}

Thus, the magnitudes of the scalar and vector products are 10 and \sqrt{2} respectively.

Therefore, the correct option is (3).