Given B = ((3 5) (2 -3) (2 0)) and C = ((-4 5) (2 0)) Find B^T B + 3C. Find the angle between the vectors a = 6i + 2j + k and b = -2i + 3j - k.

Steps to Solve

1. Matrix Transpose of B

First, we calculate the transpose of matrix $B$:

$$ B^T = \begin{pmatrix} 3 & 2 & 0 \ 5 & -3 & 2 \end{pmatrix} $$

2. Matrix Multiplication of $B^T$ and $B$

Next, we multiply $B^T$ by $B$:

$$ B^T B = \begin{pmatrix} 3 & 2 & 0 \ 5 & -3 & 2 \end{pmatrix} \begin{pmatrix} 3 & 5 \ 2 & -3 \ 0 & 2 \end{pmatrix} $$

Calculating the product step by step:

• First row, first column: $33 + 22 + 0*0 = 9 + 4 + 0 = 13$
• First row, second column: $35 + 2(-3) + 0*2 = 15 - 6 + 0 = 9$
• Second row, first column: $5*3 + (-3)2 + 20 = 15 - 6 + 0 = 9$
• Second row, second column: $55 + (-3)(-3) + 2*2 = 25 + 9 + 4 = 38$

Thus,

$$ B^T B = \begin{pmatrix} 13 & 9 \ 9 & 38 \end{pmatrix} $$

3. Calculate $3C$

Now we compute $3C$ where $C = \begin{pmatrix} -4 & 5 \ 2 & 0 \end{pmatrix}$:

$$ 3C = 3 \begin{pmatrix} -4 & 5 \ 2 & 0 \end{pmatrix} = \begin{pmatrix} -12 & 15 \ 6 & 0 \end{pmatrix} $$

4. Calculate $B^T B + 3C$

Finally, we add $B^T B$ and $3C$ together:

$$ B^T B + 3C = \begin{pmatrix} 13 & 9 \ 9 & 38 \end{pmatrix} + \begin{pmatrix} -12 & 15 \ 6 & 0 \end{pmatrix} $$

Calculating element-wise:

• First row, first column: $13 + (-12) = 1$
• First row, second column: $9 + 15 = 24$
• Second row, first column: $9 + 6 = 15$
• Second row, second column: $38 + 0 = 38$

So,

$$ B^T B + 3C = \begin{pmatrix} 1 & 24 \ 15 & 38 \end{pmatrix} $$

5. Find the Angle Between Vectors a and b

Using the vectors $a = 6i + 2j + k$ and $b = -2i + 3j - k$, we find the angle using the cosine formula:

$$ \cos(\theta) = \frac{a \cdot b}{|a| |b|} $$

6. Calculate $a \cdot b$

Calculating the dot product:

$$ a \cdot b = (6)(-2) + (2)(3) + (1)(-1) = -12 + 6 - 1 = -7 $$

7. Calculate magnitudes $|a|$ and $|b|$

For vector $a$:

$$ |a| = \sqrt{6^2 + 2^2 + 1^2} = \sqrt{36 + 4 + 1} = \sqrt{41} $$

For vector $b$:

$$ |b| = \sqrt{(-2)^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} $$

8. Calculate the angle $\theta$

Using the values in the cosine formula:

$$ \cos(\theta) = \frac{-7}{\sqrt{41} \sqrt{14}} $$

Finally, use the inverse cosine to find $\theta$:

$$ \theta = \cos^{-1}\left(\frac{-7}{\sqrt{41 \cdot 14}} \right) $$