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Chapter 2: Problem 267

\(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\) and \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{~A}^{-}+\mathrm{B}^{-}\right|\) What is the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) \((\mathrm{A})(\pi / 6)\) (B) \((5 \pi / 6)\) (C) \((2 \pi / 3)\) (D) \((\pi / 3)\)

Short Answer

Expert verified
The angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) is \(\boxed{(D) \frac{\pi}{3}}\).

Step by step solution

01

Find the dot product

Since \(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\), their dot product is equal to 0. Thus, we can write: \((\mathrm{A}^{-}+\mathrm{B}^{-})\cdot \mathrm{A}^{-}=0\)
02

Expand the dot product

We can expand the dot product as: \(\mathrm{A}^{-}\cdot\mathrm{A}^{-}+\mathrm{B}^{-}\cdot\mathrm{A}^{-}=0\)
03

Use the magnitude relationship

Now, we use the relationship between the magnitudes of \(\mathrm{B}^{-}\) and \(\mathrm{A}^{-}+\mathrm{B}^{-}\) which is \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{A}^{-}+\mathrm{B}^{-}\right|\). Square both sides: \begin{align*} \left|\mathrm{B}^{-}\right|^2 &= 4\left|\mathrm{A}^{-}+\mathrm{B}^{-}\right|^2 \\ \left(\mathrm{B}^{-}\cdot\mathrm{B}^{-}\right) &= 4\left(\mathrm{A}^{-}+\mathrm{B}^{-}\right)\cdot\left(\mathrm{A}^{-}+\mathrm{B}^{-}\right) \end{align*}
04

Expand and simplify

Expand and simplify the equation above: \begin{align*} \mathrm{B}^{-}\cdot\mathrm{B}^{-}&=4\left(\mathrm{A}^{-}\cdot\mathrm{A}^{-}+2\mathrm{A}^{-}\cdot\mathrm{B}^{-}+\mathrm{B}^{-}\cdot\mathrm{B}^{-}\right) \\ \mathrm{B}^{-}\cdot\mathrm{B}^{-}&=4\mathrm{A}^{-}\cdot\mathrm{A}^{-}+8\mathrm{A}^{-}\cdot\mathrm{B}^{-}+4\mathrm{B}^{-}\cdot\mathrm{B}^{-} \\ 3\mathrm{B}^{-}\cdot\mathrm{B}^{-}&=4\mathrm{A}^{-}\cdot\mathrm{A}^{-}+8\mathrm{A}^{-}\cdot\mathrm{B}^{-} \end{align*}
05

Dot product of B with A

We can express the dot product of \(\mathrm{B}^{-}\) with \(\mathrm{A}^{-}\) in terms of the previous result: \(\mathrm{B}^{-}\cdot\mathrm{A}^{-}=-\frac{1}{2}\mathrm{A}^{-}\cdot\mathrm{A}^{-}\)
06

Cosine of the angle between A and B

Now, let \(\theta\) be the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) (not \(\mathrm{B}^{-}\)). Recall that the dot product can be expressed in terms of the angle between two vectors and their magnitudes. So: \begin{align*} \mathrm{A}^{-}\cdot\mathrm{B}^{-}&=\left|\mathrm{A}^{-}\right|\left|\mathrm{B}^{\rightarrow}\right|\cos(\pi-\theta) \\ -\frac{1}{2}\mathrm{A}^{-}\cdot\mathrm{A}^{-}&=\left|\mathrm{A}^{-}\right|\left|\mathrm{B}^{-}\right|\cos(\pi-\theta) \end{align*}
07

Solve for the angle

Finally, we solve for the angle \(\theta\) by dividing both sides by \(\left|\mathrm{A}^{-}\right|\left|\mathrm{B}^{-}\right|\). Recall that \(\cos(\pi-\theta)=-\cos(\theta)\). So: \begin{align*} \cos(\theta)&=\frac{1}{2} \\ \theta&=\frac{\pi}{3} \end{align*} Thus, the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) is \(\boxed{(D) \frac{\pi}{3}}\).

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Most popular questions from this chapter

Which from the following is true? (A) $\cos \theta=\left[\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right) / \mathrm{AB}\right]$ (B) $\sin \theta=\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]$ (C) $\tan \theta=\left[\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right) /\left(\mathrm{A}^{-}-\mathrm{B}^{-}\right)\right]$ (D) $\cot \theta=\left[\mathrm{AB} /\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right)\right]$

To introduce a vector quantity .... (A) it needs magnitude not direction (B) it needs direction not magnitude (C) it needs both magnitude and direction (D) nothing is needed

\(\mathrm{A}=+\mathrm{i} \wedge+\mathrm{j} \wedge-2 \mathrm{k} \wedge\) 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) \([1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 j \wedge-3 \mathrm{k} \wedge)\)

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) \(\mathrm{V}_{0} \sin (\theta / 2)\)

An object moves in a straight line. It starts from the rest and its acceleration is \(2 \mathrm{~ms}^{-2}\). After reaching a certain point it comes back to the original point. In this movement its acceleration is $-3 \mathrm{~ms}^{-2}$. till it comes to rest. The total time taken for the movement is 5 second. Calculate the maximum velocity. (A) \(6 \mathrm{~ms}^{-1}\) (B) \(5 \mathrm{~ms}^{-1}\) (C) \(10 \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)
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