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Given the vectors \(\mathbf{M}=-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z}\) and \(\mathbf{N}=8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}\), find: ( \(a\) ) a unit vector in the direction of \(-\mathbf{M}+2 \mathbf{N} ;(b)\) the magnitude of \(5 \mathbf{a}_{x}+\) \(\mathbf{N}-3 \mathbf{M} ;(c)|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\)

Short Answer

Expert verified
1. The unit vector in the direction of \(-\mathbf{M}+2 \mathbf{N}\) is \(\frac{26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}}{\sqrt{756}}\). 2. The magnitude of \(5 \mathbf{a}_{x} + \mathbf{N} - 3 \mathbf{M}\) is \(\sqrt{2138}\). 3. The result of \(|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\) is \(-272 \mathbf{a}_{x}+1496 \mathbf{a}_{y}-1360 \mathbf{a}_{z}\).

Step by step solution

01

Calculate \(-\mathbf{M}+2 \mathbf{N}\) Vector#

First, let's perform the vector arithmetic to find \(-\mathbf{M}+2 \mathbf{N}\): \begin{align*} -\mathbf{M}+2 \mathbf{N} &= -(-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z})+2(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}) \\ &= 10 \mathbf{a}_{x}-4 \mathbf{a}_{y}+8 \mathbf{a}_{z}+16 \mathbf{a}_{x}+14 \mathbf{a}_{y}-4 \mathbf{a}_{z} \\ &= (10+16) \mathbf{a}_{x}+(-4+14) \mathbf{a}_{y}+(8-4) \mathbf{a}_{z} \\ &= 26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z} \end{align*}The resulting vector is \(26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}\).
02

Calculate Unit Vector in the Direction of \(-\mathbf{M}+2 \mathbf{N}\)

To find the unit vector, we need to divide the components of the resulting vector by its magnitude. First, find the magnitude: $$|\mathbf{V}|=\sqrt{(26)^2+(10)^2+(4)^2}=\sqrt{26^2 + 10^2 + 4^2} = \sqrt{756}$$Now, find the unit vector:$$\hat{\mathbf{V}} = \frac{26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}}{\sqrt{756}}$$
03

Calculate \(5 \mathbf{a}_{x}+\mathbf{N}-3\mathbf{M}\) Vector#

Next, let's perform the vector arithmetic to find \(5 \mathbf{a}_{x}+\mathbf{N}-3\mathbf{M}\): \begin{align*} 5 \mathbf{a}_{x}+\mathbf{N}-3\mathbf{M}&=5 \mathbf{a}_{x}+(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}) -3(-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z}) \\ &= 5 \mathbf{a}_{x}+8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}+30 \mathbf{a}_{x}-12 \mathbf{a}_{y}+24 \mathbf{a}_{z} \\ &= (5+8+30) \mathbf{a}_{x}+(7-12) \mathbf{a}_{y}+(-2+24) \mathbf{a}_{z} \\ &= 43 \mathbf{a}_{x}-5 \mathbf{a}_{y}+22 \mathbf{a}_{z} \end{align*}The resulting vector is \(43 \mathbf{a}_{x}-5 \mathbf{a}_{y}+22 \mathbf{a}_{z}\).
04

Calculate Magnitude of \(5 \mathbf{a}_{x}+\mathbf{N}-3\mathbf{M}\)#

Find the magnitude of the resulting vector: $$|\mathbf{W}|=\sqrt{(43)^2+(-5)^2+(22)^2}=\sqrt{43^2 + 5^2 + 22^2} = \sqrt{2138}$$
05

Calculate \(|\mathbf{M}||2\mathbf{N}|(\mathbf{M}+\mathbf{N})\)#

To solve this, first find the magnitude of \(\mathbf{M}\) and \(2\mathbf{N}\): \begin{align*} |\mathbf{M}|=\sqrt{(-10)^2+(4)^2+(-8)^2}=\sqrt{10^2+4^2+8^2}=\sqrt{136} \\ 2\mathbf{N}=2(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z})=16 \mathbf{a}_{x}+14 \mathbf{a}_{y}-4 \mathbf{a}_{z}\\ |2\mathbf{N}|=\sqrt{(16)^2+(14)^2+(-4)^2}=\sqrt{16^2+14^2+4^2}=\sqrt{372} \end{align*}Now, find the vector \(\mathbf{M}+\mathbf{N}\): \begin{align*} \mathbf{M}+\mathbf{N}&=(-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8\mathbf{a}_{z})+(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}) \\ &= (-10+8) \mathbf{a}_{x}+(4+7) \mathbf{a}_{y}+(-8-2) \mathbf{a}_{z} \\ &= -2 \mathbf{a}_{x}+11 \mathbf{a}_{y}-10 \mathbf{a}_{z} \end{align*}Finally, calculate \(|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\):$$\sqrt{136} \cdot \sqrt{372}(-2 \mathbf{a}_{x}+11 \mathbf{a}_{y}-10 \mathbf{a}_{z})=-272 \mathbf{a}_{x}+1496 \mathbf{a}_{y}-1360 \mathbf{a}_{z}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnitude Calculation
Calculating the magnitude of a vector is like measuring its length in space. To find the magnitude of a vector, you use the formula
  • \[ | extbf{V}| = \sqrt{x^2 + y^2 + z^2} \]
where \(x\), \(y\), and \(z\) are the components of the vector. For instance, with \( extbf{M} = -10 \textbf{a}_x + 4 \textbf{a}_y - 8 \textbf{a}_z\), the magnitude is
  • \[ | extbf{M}| = \sqrt{(-10)^2 + 4^2 + (-8)^2} = \sqrt{136} \]
It helps you understand how large the vector is without considering its direction. This technique is used often in physics to evaluate forces or velocities.
Unit Vector
A unit vector indicates direction but has a magnitude of 1. To find a unit vector in the direction of a given vector \(\textbf{V} = x \textbf{a}_x + y \textbf{a}_y + z \textbf{a}_z\), you divide each component by the vector's magnitude:
  • \[ \hat{\textbf{V}} = \frac{x \textbf{a}_x + y \textbf{a}_y + z \textbf{a}_z}{|\textbf{V}|} \]
For example, to find the unit vector in the direction of \(-\textbf{M} + 2\textbf{N} = 26 \textbf{a}_x + 10 \textbf{a}_y + 4 \textbf{a}_z\), you calculate the magnitude first, \(\sqrt{756}\), and then
  • \[ \hat{\textbf{V}} = \frac{26 \textbf{a}_x + 10 \textbf{a}_y + 4 \textbf{a}_z}{\sqrt{756}} \]
Unit vectors are crucial in physics for describing directions without altering magnitudes, such as indicating the alignment of forces.
Vector Addition
Vector addition is combining two or more vectors to determine a resultant vector. Adding vectors involves summing up their components:
  • \[ \textbf{A} + \textbf{B} = (x_1 + x_2) \textbf{a}_x + (y_1 + y_2) \textbf{a}_y + (z_1 + z_2) \textbf{a}_z \]
In the exercise, \(\textbf{M} = -10 \textbf{a}_x + 4 \textbf{a}_y - 8 \textbf{a}_z\) combined with \(\textbf{N} = 8 \textbf{a}_x + 7 \textbf{a}_y - 2 \textbf{a}_z\) results in
  • \[ \textbf{M} + \textbf{N} = -2 \textbf{a}_x + 11 \textbf{a}_y - 10 \textbf{a}_z \]
This technique is particularly useful in navigation and determining net displacement in physics and is often visualized by connecting vectors head-to-tail.
Vector Multiplication
Vector multiplication generally refers to either dot product or cross product. In this context, multiplying vectors by scalars involves scaling their magnitude without altering direction. Scalar multiplication of a vector \(k \textbf{V}\) is performed by multiplying each component by the scalar:
  • \[ k \textbf{V} = (k \times x) \textbf{a}_x + (k \times y) \textbf{a}_y + (k \times z) \textbf{a}_z \]
For instance, multiplying \(2 \textbf{N}\) gives
  • \[ 2 \textbf{N} = 16 \textbf{a}_x + 14 \textbf{a}_y - 4 \textbf{a}_z \]
This concept is critical for scaling forces or other vector quantities and often precedes operations like dot or cross product.

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Most popular questions from this chapter

Two unit vectors, \(\mathbf{a}_{1}\) and \(\mathbf{a}_{2}\), lie in the \(x y\) plane and pass through the origin. They make angles \(\phi_{1}\) and \(\phi_{2}\), respectively, with the \(x\) axis \((a)\) Express each vector in rectangular components; \((b)\) take the dot product and verify the trigonometric identity, \(\cos \left(\phi_{1}-\phi_{2}\right)=\cos \phi_{1} \cos \phi_{2}+\sin \phi_{1} \sin \phi_{2} ;(c)\) take the cross product and verify the trigonometric identity \(\sin \left(\phi_{2}-\phi_{1}\right)=\sin \phi_{2} \cos \phi_{1}-\cos \phi_{2} \sin \phi_{1}\)

Express the uniform vector field \(\mathbf{F}=5 \mathbf{a}_{x}\) in \((a)\) cylindrical components; (b) spherical components.

Vector A extends from the origin to \((1,2,3)\), and vector \(\mathbf{B}\) extends from the origin to \((2,3,-2)\). Find \((a)\) the unit vector in the direction of \((\mathbf{A}-\mathbf{B})\); (b) the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of \(\mathbf{A}\) and \(\mathbf{B}\).

A vector field is specified as \(\mathbf{G}=24 x y \mathbf{a}_{x}+12\left(x^{2}+2\right) \mathbf{a}_{y}+18 z^{2} \mathbf{a}_{z} .\) Given two points, \(P(1,2,-1)\) and \(Q(-2,1,3)\), find \((a) \mathbf{G}\) at \(P ;(b)\) a unit vector in the direction of \(\mathbf{G}\) at \(Q ;(c)\) a unit vector directed from \(Q\) toward \(P ;(d)\) the equation of the surface on which \(|\mathbf{G}|=60\).

Point \(A(-4,2,5)\) and the two vectors, \(\mathbf{R}_{A M}=(20,18-10)\) and \(\mathbf{R}_{A N}=(-10,8,15)\), define a triangle. Find \((a)\) a unit vector perpendicular to the triangle; \((b)\) a unit vector in the plane of the triangle and perpendicular to \(\mathbf{R}_{A N} ;(c)\) a unit vector in the plane of the triangle that bisects the interior angle at \(A\).

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