Question

If \(\boldsymbol{a}=3 \boldsymbol{i}+\boldsymbol{j}-2 \boldsymbol{k}\) and \(\boldsymbol{b}=13 \boldsymbol{i}-\boldsymbol{j}-\boldsymbol{k}\) find (a) \(a+b\) (b) \(b-3 a\) (c) \(|\boldsymbol{b}|\) (e) \(|\boldsymbol{b}-\boldsymbol{a}|\) (d) \(\hat{\boldsymbol{a}}\)

Short answer

Answer: The sum of vectors a and b is \(16\boldsymbol{i} + 0\boldsymbol{j} - 3\boldsymbol{k}\).

Step-by-step solution

(a) Adding Vectors a and b

To add two vectors, we simply add their corresponding components. So for a + b: \((a+b)_x = a_x + b_x\) \((a+b)_y = a_y + b_y\) \((a+b)_z = a_z + b_z\) Substitute the given components of a and b in the equations above: \((a+b)_x = 3 + 13 = 16\) \((a+b)_y = 1 - 1 = 0\) \((a+b)_z = -2 - 1 = -3\) So, \(a+b = 16\boldsymbol{i} + 0\boldsymbol{j} - 3\boldsymbol{k}\)

(b) Finding \((b-3a)\)

To find \((b-3a)\), we first multiply the vector a by 3, and then subtract the components of the resulting vector from vector b. So, \((b-3a)_x = b_x - 3a_x\) \((b-3a)_y = b_y - 3a_y\) \((b-3a)_z = b_z - 3a_z\) Substitute the given components of a and b in the equations above: \((b-3a)_x = 13 - (3\times3) = 13 - 9 = 4\) \((b-3a)_y = -1 - (3\times 1) = -1 - 3 = -4\) \((b-3a)_z = -1 - (3\times -2) = -1 + 6 = 5\) So, \(b-3a = 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k}\)

(c) Finding the Magnitude of Vector b

To find the magnitude of a vector, we use the formula \(|b| = \sqrt{b_x^2 + b_y^2 + b_z^2}\). Substitute the components of vector b in the formula: \(|b| = \sqrt{(13^2) + (-1)^2 + (-1)^2 } = \sqrt{169 + 1 + 1} = \sqrt{171}\) So, \(|\boldsymbol{b}| = \sqrt{171}\)

(d) Finding the Unit Vector of Vector a

To find the unit vector of a vector, we divide the vector by its magnitude. First find the magnitude of a: \(|a| = \sqrt{\left(3^2\right) + \left(1^2\right) + \left(-2^2\right)} = \sqrt{14}\) Now, divide vector a by its magnitude: \(\hat{\boldsymbol{a}} = \frac{\boldsymbol{a}}{|\boldsymbol{a}|} = \frac{1}{\sqrt{14}}(3\boldsymbol{i}+\boldsymbol{j}-2\boldsymbol{k})\) So, \(\hat{\boldsymbol{a}} = \frac{3}{\sqrt{14}}\boldsymbol{i} + \frac{1}{\sqrt{14}}\boldsymbol{j} - \frac{2}{\sqrt{14}}\boldsymbol{k}\)

(e) Finding the Magnitude of the Difference between Vectors b and a

To find the magnitude of the difference between two vectors, we first find the difference vector and then find its magnitude using the formula \(|\boldsymbol{c}| = \sqrt{c_x^2 + c_y^2 + c_z^2}\), where c = b - a. First, find the difference vector: \((b-a)_x = b_x - a_x = 13 - 3 = 10\) \((b-a)_y = b_y - a_y = -1 - 1 = -2\) \((b-a)_z = b_z - a_z = -1 - (-2) = -1 + 2 = 1\) So, \(b-a = 10\boldsymbol{i} - 2\boldsymbol{j} + \boldsymbol{k}\) Now, find the magnitude: \(|b-a| = \sqrt{(10^2) + (-2)^2 + (1)^2} = \sqrt{100 + 4 + 1} = \sqrt{105}\) So, \(|\boldsymbol{b}-\boldsymbol{a}| = \sqrt{105}\)

Key concepts

Vector Addition

In vector addition, you combine two or more vectors to find a resultant vector. This is done by adding corresponding components of the vectors together. For example, if you have vectors \( \boldsymbol{a} = 3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k} \) and \( \boldsymbol{b} = 13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k} \), their sum is calculated by adding the \( i, j, \) and \( k \) components separately.
To find \( \boldsymbol{a} + \boldsymbol{b} \):

• Add the \( i \)-components: \( 3 + 13 = 16 \).
• Add the \( j \)-components: \( 1 - 1 = 0 \).
• Add the \( k \)-components: \( -2 - 1 = -3 \).

Therefore, the resultant vector \( \boldsymbol{a} + \boldsymbol{b} \) is \( 16\boldsymbol{i} + 0\boldsymbol{j} - 3\boldsymbol{k} \). Vector addition is a straightforward process that involves simple arithmetic with each corresponding component. This operation is essential in physics and engineering to determine resultant forces and velocities.

Vector Subtraction

Vector subtraction is similar to addition, but instead of adding, you subtract the components of one vector from the other. This operation finds the difference between two vectors and is crucial for finding relative movements or changes.
Given vectors \( \boldsymbol{b} = 13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k} \) and \( 3\boldsymbol{a} = 9\boldsymbol{i} + 3\boldsymbol{j} - 6\boldsymbol{k} \), to compute \( \boldsymbol{b} - 3\boldsymbol{a} \), do the following:

• Subtract the \( i \)-components: \( 13 - 9 = 4 \).
• Subtract the \( j \)-components: \( -1 - 3 = -4 \).
• Subtract the \( k \)-components: \( -1 + 6 = 5 \).

Thus, \( \boldsymbol{b} - 3\boldsymbol{a} = 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k} \). This method allows you to determine how one vector acts relative to another, which is a vital skill in physics when analyzing forces or velocities.

Magnitude of a Vector

The magnitude of a vector is a measure of its length. It's a scalar value and is calculated using the Pythagorean theorem extended into three dimensions. Suppose you have a vector \( \boldsymbol{b} = 13\boldsymbol{i} - \boldsymbol{j} - \boldsymbol{k} \). Its magnitude \( |\boldsymbol{b}| \) is found as follows:

• Calculate the square of each component: \( 13^2 = 169, (-1)^2 = 1, (-1)^2 = 1 \).
• Add the squares: \( 169 + 1 + 1 = 171 \).
• Take the square root: \( \sqrt{171} \).

Hence, \( |\boldsymbol{b}| = \sqrt{171} \). The magnitude gives you a sense of the size of the vector irrespective of its direction. It is extensively utilized in physics to compute distances or speeds.

Unit Vector

A unit vector has a magnitude of 1 and points in the direction of the original vector. To find the unit vector \( \hat{\boldsymbol{a}} \) of \( \boldsymbol{a} = 3\boldsymbol{i} + \boldsymbol{j} - 2\boldsymbol{k} \), first find the magnitude \( |\boldsymbol{a}| \).
Calculate \( |\boldsymbol{a}| \):

• Square each component: \( 3^2 = 9, 1^2 = 1, (-2)^2 = 4 \).
• Add these squares: \( 9 + 1 + 4 = 14 \).
• Take the square root: \( \sqrt{14} \).

Now divide each component of \( \boldsymbol{a} \) by \( \sqrt{14} \):

• \( \frac{3}{\sqrt{14}}\boldsymbol{i} \)
• \( \frac{1}{\sqrt{14}}\boldsymbol{j} \)
• \( \frac{-2}{\sqrt{14}}\boldsymbol{k} \)

So, \( \hat{\boldsymbol{a}} = \frac{3}{\sqrt{14}}\boldsymbol{i} + \frac{1}{\sqrt{14}}\boldsymbol{j} - \frac{2}{\sqrt{14}}\boldsymbol{k} \). The concept of a unit vector is crucial as it helps signify direction without regard to magnitude, aiding in various applications such as normalizing vectors for unit direction.