Question

Let a =(1/2√2, √3/2√2, 1/√2) and b=(1/√2, 0, 1√2)

i) Find the direct cosines of a and b.
ii) Find the angle between a and b.

Answer 1

Answer on Question #44583 - Math - Linear Algebra

Question 1. Let $a = (1/(2\sqrt{2}), \sqrt{3}/(2\sqrt{2}), 1/\sqrt{2})$ and $b = (1/\sqrt{2}, 0, 1/\sqrt{2})$.

1. Find the direct cosines of $a$ and $b$;

2. Find the angle between $a$ and $b$.

Solution. (i) We have

$$\begin{array}{l} |a| = \sqrt{ \left(1 / (2\sqrt{2})\right)^2 + \left(\sqrt{3}/2\sqrt{2}\right)^2 + \left(1/\sqrt{2}\right)^2 } \\ = \sqrt{1/8 + 3/8 + 1/2} \\ = \sqrt{1} \\ = 1. \end{array}$$

Hence, the direct cosines of $a$ are

$$\begin{array}{l} \alpha_1 = \frac{1/(2\sqrt{2})}{1} = (1/(2\sqrt{2}); \\ \beta_1 = \frac{\sqrt{3}/(2\sqrt{2})}{1} = \sqrt{3}/(2\sqrt{2}); \\ \gamma_1 = \frac{1/\sqrt{2}}{1} = 1/\sqrt{2}. \end{array}$$

Similarly

$$\begin{array}{l} |b| = \sqrt{ \left(1 / \sqrt{2}\right)^2 + \left(0^2 + \left(1/\sqrt{2}\right)^2\right) } \\ = \sqrt{1/2 + 0 + 1/2} \\ = \sqrt{1} \\ = 1. \end{array}$$

Hence, the direct cosines of $b$ are

$$\begin{array}{l} \alpha_2 = \frac{1/\sqrt{2}}{1} = 1/\sqrt{2}; \\ \beta_2 = \frac{0}{1} = 0; \\ \gamma_2 = \frac{1/\sqrt{2}}{1} = 1/\sqrt{2}. \end{array}$$

(ii) Let $\theta$ denote the angle between $a$ and $b$. Then

$$\begin{array}{l} \cos \theta = \alpha_ {1} \alpha_ {2} + \beta_ {1} \beta_ {2} + \gamma_ {1} \gamma_ {2} \\ = (1 / (2 \sqrt {2}) (1 / \sqrt {2}) + (\sqrt {3} / (2 \sqrt {2})) 0 + (1 / \sqrt {2}) (1 / \sqrt {2}) \\ = 1 / 4 + 0 + 1 / 2 \\ = 3 / 4. \end{array}$$

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