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NEET 2022
Exam
NEET
Year
2022
Subject
Physics
Chapter
Mathematics in Physics

NEET 2022 · Physics · Mathematics in Physics

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

If F→=2i^+j^−k^\overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k} and r⃗=3i^+2j^−2k^\vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k}, then the scalar and vector products of F⃗\vec{F} and r⃗\vec{r} have the magnitudes respectively as:

Options

  1. A5,35, \sqrt{3}
  2. B4,54, \sqrt{5}
  3. C10,210, \sqrt{2}
  4. D10, 2

Correct answer

C. 10,210, \sqrt{2}

Step-by-step solution

F→=2i^+j^−k^r⃗=3i^+2j^−2k^ (given) \begin{aligned} & \overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k} \\ & \vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k} \text{ (given) } \end{aligned} The scalar product of F→andr⃗\overrightarrow{\mathrm{F}} and \vec{r} is F⃗.r⃗=(2)(3)+(1)(2)+(−1)(−2)=6+2+2=10\begin{aligned} \vec{F} . \vec{r} & =(2)(3)+(1)(2)+(-1)(-2) \\ & =6+2+2=10 \end{aligned} The vector product of F⃗andr⃗\vec{F} and \vec{r} is F→×r⃗=∣i^j^k^21−132−2∣=i^[(1×−2)−(2×−1)]−j^[(2×−2)−(−1×3)]+k^[(2×2)−(3×1)]=i^(0)−j^(−1)+k^(1)=j^+k^\begin{aligned} \overrightarrow{\mathrm{F}} \times \vec{r} & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 3 & 2 & -2 \end{array}\right| \\ & =\hat{i}[(1 \times-2)-(2 \times-1)]-\hat{j}[(2 \times-2) \\ & -(-1 \times 3)]+\hat{k}[(2 \times 2)-(3 \times 1)] \\ & =\hat{i}(0)-\hat{j}(-1)+\hat{k}(1)=\hat{j}+\hat{k} \end{aligned} Magnitude =(1)2+(1)2=2= \sqrt{(1)^2+(1)^2}=\sqrt{2}
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