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Chapter 14: Problem 12

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

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

Step by step solution

01

(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, \(a+b = 16\boldsymbol{i} + 0\boldsymbol{j} - 3\boldsymbol{k}\)
02

(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, \(b-3a = 4\boldsymbol{i} - 4\boldsymbol{j} + 5\boldsymbol{k}\)
03

(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, \(|\boldsymbol{b}| = \sqrt{171}\)
04

(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, \(\hat{\boldsymbol{a}} = \frac{3}{\sqrt{14}}\boldsymbol{i} + \frac{1}{\sqrt{14}}\boldsymbol{j} - \frac{2}{\sqrt{14}}\boldsymbol{k}\)
05

(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, \(|\boldsymbol{b}-\boldsymbol{a}| = \sqrt{105}\)

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Most popular questions from this chapter

Given three vectors \(\boldsymbol{a}, \boldsymbol{b}\) and \(\boldsymbol{c}\), their triple scalar product is defined to be \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\). It can be shown that the modulus of this is the volume of the parallelepiped formed by the three vectors. Find the volume of the parallelepiped formed by the three vectors \(\boldsymbol{a}=3 \boldsymbol{i}+\boldsymbol{j}-2 \boldsymbol{k}, \boldsymbol{b}=\boldsymbol{i}+2 \boldsymbol{j}-2 \boldsymbol{k}\) and \(\boldsymbol{c}=2 \boldsymbol{i}+5 \boldsymbol{j}+\boldsymbol{k}\)

Write down the vector equation of the line passing through the points with position vectors $$ \begin{aligned} &p=3 i+7 j-2 k \\ &q=-3 i+2 j+2 k \end{aligned} $$ Find also the cartesian equation of this line.

State, or find out, which of the following are scalars and which are vectors: (a) the volume of a petrol tank, (b) a length measured in metres, (c) a length measured in miles, (d) the angular velocity of a flywheel, (e) the relative velocity of two aircraft, (f) the work done by a force, (g) electrostatic potential, (h) the momentum of an atomic particle.

Given that \(p=2 i+2 j\) and \(q=7 k\) find \(p \cdot q\) and interpret this result geometrically.

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.
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