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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 $\vec{b}$?
Solution :
Step 1: Analyzing the Dot Product Constraint
Given the scalar (dot) product: $\vec{a} \cdot \vec{b} = 0$ [Per the definition of the dot product: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ is the angle between the vectors].
This equation holds true if and only if:
- $|\vec{a}| = 0$ (i.e., $\vec{a} = \vec{0}$), or
- $|\vec{b}| = 0$ (i.e., $\vec{b} = \vec{0}$), or
- $\cos \theta = 0 \implies \theta = 90^\circ$ ($\vec{a} \perp \vec{b}$).
Step 2: Analyzing the Cross Product Constraint
Given the vector (cross) product: $\vec{a} \times \vec{b} = \vec{0}$ [Per the definition of the cross product: $\vec{a} \times \vec{b} = (|\vec{a}| |\vec{b}| \sin \theta) \hat{n}$].
This equation holds true if and only if:
- $|\vec{a}| = 0$ (i.e., $\vec{a} = \vec{0}$), or
- $|\vec{b}| = 0$ (i.e., $\vec{b} = \vec{0}$), or
- $\sin \theta = 0 \implies \theta = 0^\circ$ or $180^\circ$ ($\vec{a} \parallel \vec{b}$).
Step 3: Synthesis and Logical Conclusion
We are given that both conditions must be satisfied simultaneously:
| Condition | Geometric Implication |
|---|---|
| $\vec{a} \cdot \vec{b} = 0$ | $\vec{a} = \vec{0}$, $\vec{b} = \vec{0}$, or $\vec{a} \perp \vec{b}$ |
| $\vec{a} \times \vec{b} = \vec{0}$ | $\vec{a} = \vec{0}$, $\vec{b} = \vec{0}$, or $\vec{a} \parallel \vec{b}$ |
If we assume neither $\vec{a}$ nor $\vec{b}$ is a zero vector, we face a contradiction: the vectors cannot be simultaneously perpendicular ($\theta = 90^\circ$) and parallel ($\theta = 0^\circ$ or $180^\circ$).
Therefore, the only mathematically consistent conclusion is that at least one of the vectors must be a zero vector. If $\vec{a} \neq \vec{0}$, then $|\vec{a}| \neq 0$; if $\vec{b} \neq \vec{0}$, then $|\vec{b}| \neq 0$. Since $\sin \theta$ and $\cos \theta$ cannot be zero at the same value of $\theta$, it forces the magnitude of at least one vector to be zero.
Visual Representation of Logical Contradiction
Final Solution: Either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ (or both).
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 $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.
- 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 $\vec{a}$.
- Q4: Show that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$
- Q5: Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.
- 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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