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Chapter 2: Problem 288

What is unit vector along \(\hat{i}+\hat{\jmath}\) ? (A) \([(1+\hat{j}) / 2]\) (C) \([(\hat{1}+\hat{j}) / \sqrt{3}]\) (D) \([(\hat{1}-\hat{j}) / \sqrt{2}]\)

Short Answer

Expert verified
The correct answer is \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{\jmath})\). None of the provided options match this answer.

Step by step solution

01

Write down the given vector in component form

We are given the vector \(\hat{i}+\hat{\jmath}\) which in component form can be written as: \[ \vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
02

Calculate the magnitude of the vector

To find the unit vector along \(\vec{v}\), we first need to find the magnitude of \(\vec{v}\), which is the square root of the sum of the squares of the components: \[ |\vec{v}| = \sqrt{(1^2)+(1^2)} = \sqrt{2} \]
03

Normalize the vector

Now that we have the magnitude, we can normalize the vector by dividing each component by the magnitude to get the unit vector: \[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} \] The unit vector in i-j representation is: \[ \hat{u} = \frac{1}{\sqrt{2}}(\hat{i}+\hat{\jmath}) \] Comparing with the given options, we find that the correct answer is not among the options. There must be a mistake in the options provided.

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Most popular questions from this chapter

\(\mathrm{A}=+\mathrm{i} \wedge+\mathrm{j} \wedge-2 \mathrm{k} \wedge\) and $\mathrm{B} \overrightarrow{\mathrm{i}} \wedge-\mathrm{j} \wedge+\mathrm{k} \wedge$ Find the unit vector in direction of \(\mathrm{A} \rightarrow \times \mathrm{B}^{\rightarrow}\) (A) $[1 / \sqrt{(23)}](-\mathrm{i} \wedge-5 \mathrm{j} \wedge-2 \mathrm{k} \wedge)$ (B) $[1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 \mathrm{j} \wedge-3 \mathrm{k} \wedge)$ (C) \([1 / \sqrt{(29})](-i \wedge-5 j \wedge-3 k \wedge)\) (D) \([1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 j \wedge-3 \mathrm{k} \wedge)\)

Equation of a projectile is given by \(\mathrm{y}=\mathrm{Ax}-\mathrm{Bx}^{2}\). Find the range for the particle. (A) \((\mathrm{A} / \mathrm{B})\) (B) \((\mathrm{A} / 4 \mathrm{~B})\) (C) \((\mathrm{A} / 2 \mathrm{~B})\) (D) \((2 \mathrm{~A} / \mathrm{B})\)

A stone is projected with an angle \(\theta\) and velocity \(\mathrm{V}_{0}\) from point \(P\). It strikes the ground at point \(Q\). If the both \(P\) and \(Q\) are on same horizontal line, then find average velocity. (A) \(V_{0} \cos \theta\) (B) \(\mathrm{V}_{0} \sin \theta\) (C) \(\mathrm{V}_{0} \cos (\theta / 2)\) (D) \(\mathrm{V}_{0} \sin (\theta / 2)\)

If \(\mathrm{A}^{\rightarrow}=3 \hat{1}+4 \hat{\jmath}+9 \mathrm{k}\) is multiplied by 3 , then the component of the new vector along \(\mathrm{z}\) direction is .. (A) \(-3\) (B) \(+3\) \(\begin{array}{ll}\text { (C) }-27 & \text { (D) }+27\end{array}\)

A particle goes from point \(\mathrm{A}\) to \(\mathrm{B}\). Its displacement is \(\mathrm{X}\) and path length is \(\mathrm{y}\). So $\mathrm{x} / \mathrm{y} \ldots$ \((\mathrm{A})>1\) (B) \(<1\) (C) \(\geq 1\) (D) \(\leq 1\)
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