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Chapter 14: Problem 9

Find the projection of the vector \(6 \boldsymbol{i}+\boldsymbol{j}+5 \boldsymbol{k}\) onto the vector \(\boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}\).

Short Answer

Expert verified
Answer: The projection of vector A onto vector B is \(\frac{5}{2}\boldsymbol{i}-\frac{5}{2}\boldsymbol{j}+5\boldsymbol{k}\).

Step by step solution

01

Write given vectors

Write down both the given vectors: \(A = 6 \boldsymbol{i}+\boldsymbol{j}+5 \boldsymbol{k}\) \(B = \boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}\)
02

Find the dot product of the two vectors

To find the dot product of vectors A and B, we can use the formula \(A \cdot B = |A||B|cos(\theta)\), where \(|A|\) and \(|B|\) are the magnitudes of vectors A and B, and \(\theta\) is the angle between them. In this case, the dot product can be calculated as follows: \((6 \boldsymbol{i}+\boldsymbol{j}+5 \boldsymbol{k}) \cdot (\boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}) = 6\times1+1\times(-1)+5\times2 = 6-1+10= 15\)
03

Find the magnitude of the vector being projected upon

Calculate the magnitude of vector B: \(|B|=\sqrt{(1^2+(-1)^2+2^2)}=\sqrt{6}\)
04

Calculate the scalar projection of vector A onto vector B

The scalar projection can be calculated using the formula: \(ScalarProjection(AontoB)=\frac{A \cdot B}{|B|^2}\) So, the scalar projection of A onto B is: \(\frac{15}{(\sqrt{6})^2} = \frac{15}{6} = \frac{5}{2}\)
05

Calculate the projection vector

Now, we can calculate the projection vector by multiplying the scalar projection with the vector B: \(Projection = \frac{5}{2}(\boldsymbol{i}-\boldsymbol{j}+2\boldsymbol{k}) = \frac{5}{2}\boldsymbol{i}-\frac{5}{2}\boldsymbol{j}+\frac{5}{2}\times2\boldsymbol{k}\)
06

Write the final answer

The projection of vector A onto vector B is: \(Projection = \frac{5}{2}\boldsymbol{i}-\frac{5}{2}\boldsymbol{j}+5\boldsymbol{k}\)

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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 \(\boldsymbol{c}=2 \boldsymbol{i}+5 \boldsymbol{j}+\boldsymbol{k}\)

Two unit vectors are parallel. What can you deduce about their scalar product?

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.

Find the work done by a force of magnitude 10 newtons acting in the direction of the vector \(3 \boldsymbol{i}+\boldsymbol{j}+8 \boldsymbol{k}\) if it moves a particle from the point \((1,1,1)\) to the point \((3,1,2)\).

Find the angle between the vectors \(12 \boldsymbol{i}-\boldsymbol{j}\), and \(2 \boldsymbol{i}+\boldsymbol{j}+\boldsymbol{k}\).
See all solutions

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