Question

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

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

Write given vectors

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

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: \((6 \boldsymbol{i}+\boldsymbol{j}+5 \boldsymbol{k}) \cdot (\boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}) = 6\times1+1\times(-1)+5\times2 = 6-1+10= 15\)

Find the magnitude of the vector being projected upon

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

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: \(\frac{15}{(\sqrt{6})^2} = \frac{15}{6} = \frac{5}{2}\)

Calculate the projection vector

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

Write the final answer

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

Key concepts

Dot Product

The dot product of two vectors is a way to multiply them together, resulting in a scalar (a single number). You compute it by multiplying corresponding components of the vectors and summing them up. For vectors \( \mathbf{A} = 6 \mathbf{i} + \mathbf{j} + 5 \mathbf{k} \) and \( \mathbf{B} = \mathbf{i} - \mathbf{j} + 2 \mathbf{k} \):

• Multiply the components: \( 6 \times 1 + 1 \times (-1) + 5 \times 2 \)
• Add them up: \( 6 - 1 + 10 = 15 \)

The dot product \( (\mathbf{A} \cdot \mathbf{B}) \) helps determine the angle between vectors and is crucial for projections.

Magnitude of a Vector

The magnitude of a vector measures its length. You find it by taking the square root of the sum of the squares of its components. For vector \( \mathbf{B} = \mathbf{i} - \mathbf{j} + 2 \mathbf{k} \):

• Square each component: \( 1^2 + (-1)^2 + 2^2 = 1 + 1 + 4 \)
• Add them: \( 6 \)
• Take the square root: \( \sqrt{6} \)

The magnitude \( |\mathbf{B}| \) is needed to normalize the vector, especially when calculating projections.

Scalar Projection

Scalar projection is a measure of how much one vector extends in the direction of another. It is found by dividing the dot product of the vectors by the magnitude of the target vector squared. For vector \( \mathbf{A} \) onto \( \mathbf{B} \):

• Use the formula: \( \text{Scalar Projection} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2} \)
• Plug in the values: \( \frac{15}{6} = \frac{5}{2} \)

This gives the component of \( \mathbf{A} \) in the direction of \( \mathbf{B} \), setting the stage for calculating the projection vector.

Projection Vector

The projection vector is the vector representation of the scalar projection, extending in the direction of the target vector. It can be calculated by multiplying the scalar projection by the vector being projected onto. For \( \mathbf{A} \) onto \( \mathbf{B} \):

• Multiply scalar projection by \( \mathbf{B} \): \( \frac{5}{2}(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) \)
• Perform the multiplication: \( \frac{5}{2}\mathbf{i} - \frac{5}{2}\mathbf{j} + 5\mathbf{k} \)

The result is the projection vector, showing exactly how vector \( \mathbf{A} \) lies along vector \( \mathbf{B} \).