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Chapter 2: Problem 296

Find a unit vector from the followings. (A) \(\hat{\imath}+\hat{j}\) (B) \(\hat{\imath}-\hat{j}\) (C) \((1 / \sqrt{2}) \hat{\imath}+(1 / \sqrt{2}) \hat{\jmath}\) (D) \((1 / \sqrt{2}) \hat{1}-(1 / 2) \hat{\jmath}\)

Short Answer

Expert verified
(A) \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) (B) \(\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)\) (C) \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) (D) \(\left(\frac{(1/\sqrt{2})}{\sqrt{(1/\sqrt{2})^2 + (-1/2)^2}}, \frac{(-1/2)}{\sqrt{(1/\sqrt{2})^2 + (-1/2)^2}}\right)\)

Step by step solution

01

(A) Unit Vector

\((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\)
02

(B) Unit Vector

\((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\)
03

(C) Unit Vector

Since the magnitude is already 1, the vector itself is a unit vector: \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\)
04

(D) Unit Vector

\((\frac{(1/\sqrt{2})}{\sqrt{(1/\sqrt{2})^2 + (-1/2)^2}}, \frac{(-1/2)}{\sqrt{(1/\sqrt{2})^2 + (-1/2)^2}})\) To summarize, the unit vectors corresponding to the given vectors are: (A) \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) (B) \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\) (C) \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) (D) \((\frac{(1/\sqrt{2})}{\sqrt{(1/\sqrt{2})^2 + (-1/2)^2}}, \frac{(-1/2)}{\sqrt{(1/\sqrt{2})^2 + (-1/2)^2}})\)

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Most popular questions from this chapter

If \(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}=0\) then (A) \(\left|\mathrm{A}^{\rightarrow}\right|\) must be zero (B) \(\mathrm{B}^{\rightarrow} \mid\) must be zero (C) either \(\mathrm{A}^{\rightarrow}=0, \mathrm{~B}^{\rightarrow}=0\) or \(\theta=0\) (D) either \(\mathrm{A}^{\rightarrow}=0, \mathrm{~B}^{\rightarrow}=0\) or \(\theta=(\pi / 2)\)

Linear momentum of a particle is $(3 \mathrm{i} \wedge+2 \mathrm{j} \wedge-\mathrm{k} \wedge) \mathrm{kg} \mathrm{ms}^{-1}$. Find its magnitude. (A) \(\sqrt{14}\) (B) \(\sqrt{12}\) (C) \(\sqrt{15}\) (D) \(\sqrt{11}\)

Angle of projection, maximum height and time to reach the maximum height of a particle are \(\theta, \mathrm{H}\) and \(\mathrm{tm}\) respectively. Find the true relation. (A) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(\mathrm{H} / 2 \mathrm{~g})}\) (B) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(2 \mathrm{H} / \mathrm{g})}\) (C) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(4 \mathrm{H} / \mathrm{g})}\) (D) \(t_{\mathrm{m}}=\sqrt{(\mathrm{H} / 4 \mathrm{~g})}\)

The motion of a particle along a straight line is described by the function \(\mathrm{x}=(3 \mathrm{t}-2)^{2}\). Calculate the acceleration after $10 \mathrm{~s}$. (A) \(9 \mathrm{~ms}^{-2}\) (B) \(18 \mathrm{~ms}^{-2}\) (C) \(36 \mathrm{~ms}^{-2}\) (D) \(6 \mathrm{~ms}^{-2}\)

A particle moves with a constant acceleration $\left(2 \mathrm{~m} / \mathrm{s}^{2}\right)\( Its initial velocity is \)10 \mathrm{~m} / \mathrm{s}$. Find velocity after \(\mathrm{t}\) second. (A) \((10+\mathrm{t}) \mathrm{ms}^{-1}\) (B) \(5(2+\mathrm{t}) \mathrm{ms}^{-1}\) (C) \(2(5+\mathrm{t}) \mathrm{ms}^{-1}\) (D) \(\left(10+\mathrm{t}^{2}\right) \mathrm{ms}^{-1}\)
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