Question

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

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

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 \(26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}\).

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:$$\hat{\mathbf{V}} = \frac{26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}}{\sqrt{756}}$$

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 \(43 \mathbf{a}_{x}-5 \mathbf{a}_{y}+22 \mathbf{a}_{z}\).

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

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

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 \(|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\):$$\sqrt{136} \cdot \sqrt{372}(-2 \mathbf{a}_{x}+11 \mathbf{a}_{y}-10 \mathbf{a}_{z})=-272 \mathbf{a}_{x}+1496 \mathbf{a}_{y}-1360 \mathbf{a}_{z}$$

Key concepts

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.