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Chapter 2: Problem 287

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

Short Answer

Expert verified
\(\hat{C} = [\frac{1}{\sqrt{14}}](- \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k})\)

Step by step solution

01

Calculate the cross product of A and B

To find the cross product of A and B, we will use the following formula for cross product of two vectors: \(A \times B = (A_yB_z - A_zB_y) \mathrm{i} + (A_zB_x - A_xB_z) \mathrm{j} + (A_xB_y - A_yB_x) \mathrm{k}\) Given A = (1, 1, -2) and B = (1, -1, 1), we have \(A_x = 1, A_y = 1, A_z = -2\), and \(B_x = 1, B_y = -1, B_z = 1\). Now, substituting the values in the formula, we get: \(A \times B = ((1)(1) - (-2)(-1)) \mathrm{i} + ((-2)(1) - (1)(1)) \mathrm{j} + ((1)(-1) - (1)(1)) \mathrm{k}\)
02

Simplify the cross product expression

Now, we will simplify the cross product expression: \(A \times B = (1 - 2) \mathrm{i} + (-2 - 1) \mathrm{j} + (-1 - 1) \mathrm{k} = - \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k}\)
03

Calculate the magnitude of the cross product

To calculate the magnitude (length) of the cross product, we use the formula: \(|A \times B| = \sqrt{(\mathrm{-1})^2 + (\mathrm{-3})^2 + (\mathrm{-2})^2}\) \(|A \times B| = \sqrt{1 + 9 + 4}\) \(|A \times B| = \sqrt{14}\)
04

Calculate the unit vector in the direction of A x B

To find the unit vector, we divide each component of the cross product by its magnitude: \(\hat{C} = \frac{(- \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k})}{\sqrt{14}}\) Now, we can rewrite it as: \(\hat{C} = [\frac{1}{\sqrt{14}}](- \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k})\) Comparing this with the given options, it doesn't match any of the choices, which indicates that there may be an error in the given options. However, our calculated unit vector \(\hat{C}\) is correct.

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Most popular questions from this chapter

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 goods train is moving with constant acceleration. When engine passes through a signal its speed is U. Midpoint of the train passes the signal with speed \(\mathrm{V}\). What will be the speed of the last wagon? (B) \(\left.\sqrt{[}\left(\mathrm{V}^{2}-\mathrm{U}^{2}\right) / 2\right]\) (D) \(\sqrt{[}\left(2 \mathrm{~V}^{2}-\mathrm{U}^{2}\right)\)

Stone \(\mathrm{A}\) is thrown in horizontal direction with velocity of $10 \mathrm{~ms}^{-1}\( at the same time stone \)\mathrm{B}$ freely falls vertically in downward direction. Calculate the velocity of \(\mathrm{B}\) with respect to \(\mathrm{A}\) after 10 second. (A) \(10 \mathrm{~ms}^{-1}\) (B) \(\sqrt{(101)} \mathrm{ms}^{-1}\) (C) \(10 \sqrt{(101)} \mathrm{ms}^{-1}\) (D) 0

Velocity of particle \(\mathrm{A}\) with respect to particle \(\mathrm{B}\) is \(4(\mathrm{~m} / \mathrm{s})\) while they are moving in same direction. And it is \(10(\mathrm{~m} / \mathrm{s})\) while they are in opposite direction. What are the velocities of the particles with respect to the stationary frame of reference. (A) \(7 \mathrm{~ms}^{-1}, 3 \mathrm{~ms}^{-1}\) (B) \(4 \mathrm{~ms}^{-1}, 5 \mathrm{~ms}^{-1}\) (C) \(7 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\) (D) \(10 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\)

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) $\cot \theta=\left[\mathrm{AB} /\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right)\right]$
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