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Q11:
Show that the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} - 8\hat{k}$ are collinear.
Solution :
Step 1: Formal Definition of Collinearity
Two vectors $\vec{a}$ and $\vec{b}$ are said to be collinear if and only if there exists a non-zero scalar $\lambda \in \mathbb{R}$ such that $\vec{b} = \lambda \vec{a}$ [Per the definition of parallel/collinear vectors in vector space]. If such a scalar exists, the vectors possess the same or opposite direction and are proportional component-wise.
Step 2: Assigning Vector Variables
Let the given vectors be defined as:
- $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$
- $\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}$
Step 3: Component-wise Scalar Comparison
To determine if $\vec{b} = \lambda \vec{a}$, we equate the components:
| Component | Equation | Solving for $\lambda$ |
|---|---|---|
| $x$-component | $-4 = \lambda(2)$ | $\lambda = -2$ |
| $y$-component | $6 = \lambda(-3)$ | $\lambda = -2$ |
| $z$-component | $-8 = \lambda(4)$ | $\lambda = -2$ |
Step 4: Vector Synthesis and Logical Deduction
Since the ratio of the corresponding components is constant for all three dimensions ($x, y, z$), we can express $\vec{b}$ as follows:
$\vec{b} = -2(2\hat{i} - 3\hat{j} + 4\hat{k})$
$\vec{b} = -2\vec{a}$
By the scalar multiplication property, since $\vec{b} = \lambda \vec{a}$ where $\lambda = -2$, the vectors $\vec{a}$ and $\vec{b}$ are parallel. Because they share a line of action in vector space, they are defined as collinear [By the geometric property of parallel vectors].
Final Solution: Since $\vec{b} = -2\vec{a}$, the vectors are scalar multiples of each other, satisfying the condition for collinearity. Hence, the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} - 8\hat{k}$ are collinear.
More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.2
- Q1(i): Compute the magnitude of the following vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$
- Q1(ii): Compute the magnitude of the following vectors: $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$
- Q1(iii): Compute the magnitude of the following vectors: $\vec{c} = \frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{1}{3}\hat{k}$
- Q2: Write two different vectors having same magnitude.
- Q3: Write two different vectors having same direction.
- Q4: Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.
- Q5: Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).
- Q6: Find the sum of the vectors $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$.
- Q7: Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.
- Q8: Find the unit vector in the direction of vector $\vec{PQ}$, where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
- Q9: For given vectors, $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.
- Q10: Find a vector in the direction of vector $5\hat{i} - \hat{j} + 2\hat{k}$ which has magnitude 8 units.
- Q12: Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.
- Q13: Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1), directed from A to B.
- Q14: Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ.
- Q15(i): Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio 2 : 1 (i) internally
- Q15(ii): Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio 2 : 1 (ii) externally
- Q16: Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).
- Q17: Show that the points A, B and C with position vectors, $\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}$, respectively form the vertices of a right angled triangle.
- Q18: In triangle ABC (Fig 10.18), which of the following is not true:
- Q19: If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:
CBSE Solutions for Class 12 Mathematics Vector Algebra
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