Find the angle between two vectors a and b with magnitudes √3 and 2, respectively having a·b = √6. What is direction cosine of a line which passes through the points (2,1,2) and (4... Find the angle between two vectors a and b with magnitudes √3 and 2, respectively having a·b = √6. What is direction cosine of a line which passes through the points (2,1,2) and (4,2,0)? Read more

Steps to Solve

1. Angle Between Two Vectors

To find the angle $\theta$ between two vectors $\mathbf{a}$ and $\mathbf{b}$, we use the formula: $$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} $$ Given:

• Magnitude of $\mathbf{a}$: $|\mathbf{a}| = \sqrt{3}$
• Magnitude of $\mathbf{b}$: $|\mathbf{b}| = 2$
• Dot product $\mathbf{a} \cdot \mathbf{b} = \sqrt{6}$

Substituting these values into the equation: $$ \cos \theta = \frac{\sqrt{6}}{\sqrt{3} \cdot 2} $$

2. Calculate Cosine Value

Calculate the denominator: $$ \sqrt{3} \cdot 2 = 2\sqrt{3} $$ Now substitute: $$ \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} $$ Simplify further: $$ \cos \theta = \frac{\sqrt{6}}{\sqrt{12}} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} $$

3. Find Angle

To find $\theta$, we take the $\cos^{-1}$: $$ \theta = \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) $$ This gives: $$ \theta = \frac{\pi}{4} $$

4. Direction Cosines of the Line

To find the direction cosines for the line passing through the points $(2,1,2)$ and $(4,2,0)$, first find the direction vector $\mathbf{d}$ from these points: $$ \mathbf{d} = (4-2, 2-1, 0-2) = (2, 1, -2) $$

5. Calculate Direction Cosines

Now, the direction cosines $\ell, m, n$ are given by: $$ \ell = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} $$ where $a = 2, b = 1, c = -2$.

First, calculate the magnitude of the direction vector: $$ \sqrt{(2^2 + 1^2 + (-2)^2)} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 $$

6. Final Direction Cosines

Substituting the values: $$ \ell = \frac{2}{3}, \quad m = \frac{1}{3}, \quad n = \frac{-2}{3} $$ Thus, direction cosines are: $$ \left( \frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right) $$