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Chapter 2: Problem 278

$ \quad \mathrm{A}^{\rightarrow}=2 \mathrm{i} \wedge-3 \mathrm{j} \wedge+\mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=8 \mathrm{i} \wedge-6 \mathrm{j} \wedge-4 \mathrm{k} \wedge\( then \)\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|=\ldots$ (A) 28 (B) 14 (C) \(39.8\) (D) 7

Short Answer

Expert verified
The magnitude of the cross product of the given vectors A and B is approximately 26.93, so the closest answer choice is (A) 28.

Step by step solution

01

Calculate the cross product of A and B

Find the cross product of the two vectors A and B using the formula: \(\mathrm{A} \times \mathrm{B} = \begin{bmatrix} i & j & k \\ 2 & -3 & 1 \\ 8 & -6 & -4 \end{bmatrix}\)
02

Expand the determinant

Expand the determinant to find the cross product: \(\mathrm{A} \times \mathrm{B} = \begin{vmatrix} -3 & 1 \\ -6 & -4 \end{vmatrix}\,i - \begin{vmatrix} 2 & 1 \\ 8 & -4 \end{vmatrix}\,j + \begin{vmatrix} 2 & -3 \\ 8 & -6 \end{vmatrix}\,k\)
03

Calculate the values in the determinant

Calculate the values in the determinant to find the cross product: \(\mathrm{A} \times \mathrm{B} = [(-3 \times -4) - (1 \times -6)]\,i - [(2 \times -4) - (1 \times 8)]\,j + [(2 \times -6) - (-3 \times 8)]\,k\)
04

Further simplification

Simplify the expression to get the result of the cross product: \(\mathrm{A} \times \mathrm{B} = [(12+6)]\,i - [(-8-8)]\,j + [(-12+24)]\,k = 18i + 16j + 12k\)
05

Find the magnitude of the cross product

Calculate the magnitude of the cross product using the formula: \(|\mathrm{A} \times \mathrm{B}| = \sqrt{(18^2 + 16^2 + 12^2)}\)
06

Calculate the magnitude

Perform the calculation to find the magnitude of the cross product: \(|\mathrm{A} \times \mathrm{B}| = \sqrt{(324 + 256 + 144)} = \sqrt{724}\)
07

Check the given options

Now, look for the answer choice that is closest to \(\sqrt{724}\): (A) 28 (B) 14 (C) \(39.8\) (D) 7 Since \(|\mathrm{A} \times \mathrm{B}| = \sqrt{724} \approx 26.93\), the correct answer is (A) 28.

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Most popular questions from this chapter

A particle moves on a plane along the path \(\mathrm{y}=\mathrm{Ax}^{3}+\mathrm{B}\) in such a way that $(\mathrm{dx} / \mathrm{dt})=\mathrm{c} . \mathrm{c}, \mathrm{A}, \mathrm{B}$ are constant. Calculate the acceleration of the particle. (A) \(3 \mathrm{Ax} \mathrm{cj} \mathrm{ms}^{-2}\) (B) \(6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\) (C) \(3 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\) (D) \(\left[c \hat{i}+3 A x c^{2} \hat{j}\right] m s^{-2}\)

\(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}=\mathrm{C}^{-}\), Then \(\mathrm{C}^{\rightarrow}\) is perpendicular to (A) \(\mathrm{A}^{\rightarrow}\) only (B) \(\mathrm{B}^{\rightarrow}\) only (C) \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) both when the angle between them is \(\ldots\) (D) \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) both whatever to be the angle between them

The angle between \(i \wedge+j \wedge\) and \(z\) axis is \(\ldots\) (A) 0 (B) 45 (C) 90 (D) 180

Frame of reference is a \(\ldots\) and a ... from where an observer takes his observation, (A) place, size (B) size, situation (C) situation, size (D) place, situation

A particle moves in one direction with acceleration \(2 \mathrm{~ms}^{-2}\) and initial velocity \(3 \mathrm{~ms}^{-1}\).After what time its displacement will be \(10 \mathrm{~m}\) ? (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)
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