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Q14:
If either vector $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \cdot \vec{b} = 0$. But the converse need not be true. Justify your answer with an example.
Solution :
Step 1: Theoretical Foundation of the Scalar (Dot) Product
The scalar product of two vectors $\vec{a}$ and $\vec{b}$ is defined as: $$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$$ where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of the vectors, and $\theta$ is the angle between them [Per the definition of the inner product in Euclidean space].
Step 2: Analysis of the Direct Statement
If $\vec{a} = \vec{0}$, then $|\vec{a}| = 0$. Substituting this into the scalar product formula:
$$\vec{a} \cdot \vec{b} = 0 \cdot |\vec{b}| \cos \theta = 0$$Similarly, if $\vec{b} = \vec{0}$, $|\vec{b}| = 0$, leading to $\vec{a} \cdot \vec{b} = 0$. This confirms the initial proposition that if either vector is the null vector, their dot product is zero.
Step 3: Evaluating the Converse
The converse states: If $\vec{a} \cdot \vec{b} = 0$, then $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$. To disprove this, we seek non-zero vectors $\vec{a}$ and $\vec{b}$ such that their dot product is zero. This occurs when $\cos \theta = 0$, which happens when $\theta = 90^\circ$ ($\frac{\pi}{2}$ radians). Two non-zero vectors that are orthogonal must have a dot product of zero.
Step 4: Providing a Counter-Example
Let $\vec{a} = \hat{i} = (1, 0, 0)$ and $\vec{b} = \hat{j} = (0, 1, 0)$. Both are non-zero vectors [Magnitude $|\vec{a}| = 1 \neq 0$ and $|\vec{b}| = 1 \neq 0$].
Calculating the dot product:
$$\vec{a} \cdot \vec{b} = (1)(0) + (0)(1) + (0)(0) = 0$$Step 5: Justification Summary
| Vector | Components | Magnitude |
|---|---|---|
| $\vec{a}$ | $(1, 0, 0)$ | $1$ |
| $\vec{b}$ | $(0, 1, 0)$ | $1$ |
Since $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$, but $\vec{a} \cdot \vec{b} = 0$, the converse statement is false [By the property of orthogonal vectors in $\mathbb{R}^3$].
Final Solution: The converse is not true. For non-zero vectors $\vec{a}$ and $\vec{b}$, $\vec{a} \cdot \vec{b} = 0$ if and only if the vectors are orthogonal ($\theta = 90^\circ$). An example is $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$.
More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.3
- Q1: Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 2, respectively having $\vec{a} \cdot \vec{b} = \sqrt{6}$.
- Q2: Find the angle between the vectors $\hat{i} - 2\hat{j} + 3\hat{k}$ and $3\hat{i} - 2\hat{j} + \hat{k}$
- Q3: Find the projection of the vector $\hat{i} - \hat{j}$ on the vector $\hat{i} + \hat{j}$.
- Q4: Find the projection of the vector $\hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $7\hat{i} - \hat{j} + 8\hat{k}$.
- Q5: Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k})$, $\frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k})$, $\frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k})$ Also, show that they are mutually perpendicular to each other.
- Q6: Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.
- Q7: Evaluate the product $(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$.
- Q8: Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude and such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$.
- Q9: Find $|\vec{x}|$, if for a unit vector $\vec{a}$, $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$.
- Q10: If $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j}$ are such that $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda$.
- Q11: Show that $|\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b} - |\vec{b}|\vec{a}$, for any two nonzero vectors $\vec{a}$ and $\vec{b}$.
- Q12: If $\vec{a} \cdot \vec{a} = 0$ and $\vec{a} \cdot \vec{b} = 0$, then what can be concluded about the vector $\vec{b}$?
- Q13: If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.
- Q15: If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find $\angle ABC$. [$\angle ABC$ is the angle between the vectors $\vec{BA}$ and $\vec{BC}$].
- Q16: Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.
- Q17: Show that the vectors $2\hat{i} - \hat{j} + \hat{k}$, $\hat{i} - 3\hat{j} - 5\hat{k}$ and $3\hat{i} - 4\hat{j} - 4\hat{k}$ form the vertices of a right angled triangle.
- Q18: If $\vec{a}$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is unit vector if
CBSE Solutions for Class 12 Mathematics Vector Algebra
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