Vectors u and v are $ u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} $ and $ v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix} $ respectively. (a) Evaluate u . v. (b) Vector w makes an angle of $ \frac{\pi}{3} $ with u and $|w| = \sqrt{3}$. Calculate u . w.

Scottish Highers Maths Vectors: Vectors u and v are \( u = \be

Vectors u and v are $ u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} $ and $ v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix} $ respectively - Scottish Highers Maths - Question 5 - 2017

Question 5

$Vectors-u-and-v-are---$-u-=-\begin{pmatrix}-5-\\-1-\\--1-\end{pmatrix}-$-and--$-v-=-\begin{pmatrix}-3-\\--8-\\-6-\end{pmatrix}-$-respectively-Scottish Highers Maths-Question 5-2017.png$

Vectors u and v are $ u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} $ and $ v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix} $ respectively. (a) Evaluate u . ... show full transcript

Worked Solution & Example Answer:Vectors u and v are $ u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} $ and $ v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix} $ respectively - Scottish Highers Maths - Question 5 - 2017

Step 1

Evaluate u . v.

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Answer

To evaluate the scalar product of vectors u and v, we can use the formula:

$u . v = u_1 v_1 + u_2 v_2 + u_3 v_3$

Substituting the components from u and v:

$u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} \ \text{ and } \ v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix}$

$u . v = (5)(3) + (1)(-8) + (-1)(6)$

Calculating each term:

First term: ( 5 \times 3 = 15 )
Second term: ( 1 \times -8 = -8 )
Third term: ( -1 \times 6 = -6 )

Now, adding these results together:

$u . v = 15 - 8 - 6 = 1$

Step 2

Calculate |w|.

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Answer

We are given that the vector w makes an angle of ( \frac{\pi}{3} ) with the vector u. The scalar product can also be represented as:

$u . w = |u||w| \cos \theta$

Where:

( |u| ) is the magnitude of vector u
( |w| ) is the magnitude of vector w
( \theta ) is the angle between u and w

First, we calculate ( |u| ):

$|u| = \sqrt{5^2 + 1^2 + (-1)^2} = \sqrt{25 + 1 + 1} = \sqrt{27}$

Now we can substitute the known values into the scalar product equation:

$u . w = \sqrt{27} \cdot \sqrt{3} \cdot \cos \left(\frac{\pi}{3}\right)$

We know that ( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} ), so:

$u . w = \sqrt{27} \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{27} \cdot \sqrt{3}}{2}$

Now, simplifying further:

$u . w = \frac{\sqrt{81}}{2} = \frac{9}{2} = 4.5$

Vectors u and v are \( u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} \) and \( v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix} \) respectively - Scottish Highers Maths - Question 5 - 2017

Worked Solution & Example Answer:Vectors u and v are \( u = \begin{pmatrix} 5 \\ 1 \\ -1 \end{pmatrix} \) and \( v = \begin{pmatrix} 3 \\ -8 \\ 6 \end{pmatrix} \) respectively - Scottish Highers Maths - Question 5 - 2017

Evaluate u . v.

Calculate |w|.

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