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Q3:
If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\vec{a}$.
Solution :
Step 1: Identifying Direction Cosines
Let the unit vector $\vec{a}$ be defined in terms of its components as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$. By definition of a unit vector, the components $a_1, a_2, a_3$ correspond to the direction cosines $\cos \alpha, \cos \beta, \cos \gamma$, where $\alpha, \beta, \gamma$ are the angles made by the vector with the axes $x, y, z$ respectively.
Given:
- $\alpha = \frac{\pi}{3}$
- $\beta = \frac{\pi}{4}$
- $\gamma = \theta$ (where $\theta$ is an acute angle)
Step 2: Applying the Direction Cosine Identity
For any unit vector, the sum of the squares of its direction cosines is equal to unity: $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$ [By the property of direction cosines in 3D Euclidean space]
Substituting the known values: $$\cos^2\left(\frac{\pi}{3}\right) + \cos^2\left(\frac{\pi}{4}\right) + \cos^2 \theta = 1$$ $$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1$$ $$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$$ $$\frac{3}{4} + \cos^2 \theta = 1$$ $$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$$
Step 3: Solving for $\theta$ and Components
Since $\cos^2 \theta = \frac{1}{4}$, we have $\cos \theta = \pm \frac{1}{2}$. Given the constraint that $\theta$ is an acute angle ($0 \leq \theta \leq \frac{\pi}{2}$), $\cos \theta$ must be positive. $$\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$$
The components of the vector $\vec{a}$ are given by $(\cos \alpha, \cos \beta, \cos \gamma)$:
- $a_1 = \cos(\frac{\pi}{3}) = \frac{1}{2}$
- $a_2 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
- $a_3 = \cos(\frac{\pi}{3}) = \frac{1}{2}$
Visual Representation of Directional Angles
Final Solution: The angle $\theta = \frac{\pi}{3}$ and the components of $\vec{a}$ are $\left( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \right)$.
More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.4
- Q1: Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$.
- Q2: Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.
- Q4: Show that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$
- Q5: Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.
- Q6: Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?
- Q7: Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then show that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
- Q8: If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is the converse true? Justify your answer with an example.
- Q9: Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
- Q10: Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.
- Q11: Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is
- Q12: Area of a rectangle having vertices A, B, C and D with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is
CBSE Solutions for Class 12 Mathematics Vector Algebra
Chapters in CBSE - Class 12 Mathematics
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