If F =2 i + j - k and r =3 i +2 j -2 k , then the scalar and vector products of F and r have the magnitudes respectively as:

Correct answer: C. 10, 2 aligned & F =2 i + j - k & r =3 i +2 j -2 k (given) aligned The scalar product of F and r is aligned F . r & =(2)(3)+(1)(2)+(-1)(-2) & =6+2+2=10 aligned The vector product of F and r is aligned F × r & =| array ccc i & j & k 2 & 1 & -1 3 & 2 & -2 array | & = i [(1 ×-2)-(2 ×-1)]- j [(2 ×-2) & -(-1 × 3)]+ k [(2 × 2)-(3 × 1)] & = i (0)- j (-1)+ k (1)= j + k aligned Magnitude = (1)^2+(1)^2 = 2

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

If F =2 i + j - k and r =3 i +2 j -2 k , then the scalar and vector...

\overrightarrow{\mathrm{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:

Options

A $5, \sqrt{3}$
B $4, \sqrt{5}$
C $10, \sqrt{2}$
D10, 2

Correct answer

C. $10, \sqrt{2}$

Step-by-step solution

\begin{aligned}<br> & \overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k} \\<br> & \vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k} \text { (given) }<br> \end{aligned}

The scalar product of

\overrightarrow{\mathrm{F}}

and

\vec{r}

\begin{aligned}<br> \vec{F} . \vec{r} & =(2)(3)+(1)(2)+(-1)(-2) \\<br> & =6+2+2=10<br> \end{aligned}

The vector product of

\vec{F}

and

\vec{r}

\begin{aligned}<br> \overrightarrow{\mathrm{F}} \times \vec{r} & =\left|\begin{array}{ccc}<br> \hat{i} & \hat{j} & \hat{k} \\<br> 2 & 1 & -1 \\<br> 3 & 2 & -2<br> \end{array}\right| \\<br> & =\hat{i}[(1 \times-2)-(2 \times-1)]-\hat{j}[(2 \times-2) \\<br> & -(-1 \times 3)]+\hat{k}[(2 \times 2)-(3 \times 1)] \\<br> & =\hat{i}(0)-\hat{j}(-1)+\hat{k}(1)=\hat{j}+\hat{k}<br> \end{aligned}

Magnitude

= \sqrt{(1)^2+(1)^2}=\sqrt{2}

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