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Q19:
If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to
Solution :
Step 1: Establishing Geometric Definitions
Given two vectors $\vec{a}$ and $\vec{b}$ with an angle $\theta$ between them, we utilize the fundamental definitions of the scalar (dot) product and the vector (cross) product.
- The dot product is defined as: $|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta|$ [Per the Definition of Scalar Product].
- The magnitude of the cross product is defined as: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$, where $\theta \in [0, \pi]$ [Per the Definition of Vector Product].
Step 2: Setting up the Equivalence Equation
The problem states the condition: $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$. Substituting the definitions from Step 1:
$|\vec{a}| |\vec{b}| |\cos \theta| = |\vec{a}| |\vec{b}| \sin \theta$
Step 3: Algebraic Simplification
Assuming $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$ (as the angle $\theta$ is undefined for zero vectors), we can divide both sides by the product $|\vec{a}| |\vec{b}|$:
$|\cos \theta| = \sin \theta$
Since $\theta$ is the angle between two vectors, by definition $\theta \in [0, \pi]$. In this interval, $\sin \theta$ is always non-negative ($\sin \theta \geq 0$). Thus:
$\cos \theta = \sin \theta$ or $-\cos \theta = \sin \theta$
Step 4: Solving for $\theta$
Case 1: $\cos \theta = \sin \theta$
Dividing by $\cos \theta$ (given $\cos \theta \neq 0$):
$\tan \theta = 1$
$\theta = \frac{\pi}{4}$ or $45^\circ$
Case 2: $-\cos \theta = \sin \theta$
$\tan \theta = -1$
$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ or $135^\circ$
However, since we are dealing with the absolute value (magnitude) of the dot product, the equation $|\cos \theta| = \sin \theta$ restricts $\theta$ to the first and second quadrants where $\sin \theta$ is positive. Both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ satisfy the equality $|\cos \theta| = \sin \theta$. Specifically, for $\theta = \frac{\pi}{4}$, $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. For $\theta = \frac{3\pi}{4}$, $|\cos \frac{3\pi}{4}| = |-\frac{1}{\sqrt{2}}| = \frac{1}{\sqrt{2}}$ and $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.
Visual Representation of the Vector Relationship
Final Solution: $\theta = \frac{\pi}{4}$
More Questions from Class 12 Mathematics Vector Algebra Miscellaneous Exercise on Chapter 10
- Q1: Write down a unit vector in XY-plane, making an angle of $30^{\circ}$ with the positive direction of x-axis.
- Q2: Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.
- Q3: A girl walks 4 km towards west, then she walks 3 km in a direction $30^{\circ}$ east of north and stops. Determine the girlβs displacement from her initial point of departure.
- Q4: If $\vec{a} = \vec{b} + \vec{c}$, then is it true that $|\vec{a}| = |\vec{b}| + |\vec{c}|$? Justify your answer.
- Q5: Find the value of $x$ for which $x(\hat{i} + \hat{j} + \hat{k})$ is a unit vector.
- Q6: Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$.
- Q7: If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} - \vec{b} + 3\vec{c}$.
- Q8: Show that the points A (1, β 2, β 8), B(5, 0, β2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.
- Q9: Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $2\vec{a} + \vec{b}$ and $\vec{a} - 3\vec{b}$ externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.
- Q10: The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} + 5\hat{k}$ and $\hat{i} - 2\hat{j} - 3\hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.
- Q11: Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.
- Q12: Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c} \cdot \vec{d} = 15$.
- Q13: The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.
- Q14: If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.
- Q15: Prove that $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$, if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$.
- Q16: If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \ge 0$ only when
- Q17: Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if
- Q18: The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is
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