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Q16:
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.
Solution :
Step 1: Establishing the Position Vectors
To demonstrate the collinearity of points $A$, $B$, and $C$, we define their position vectors relative to the origin $O(0, 0, 0)$ as follows:
- $\vec{OA} = \mathbf{a} = 1\hat{i} + 2\hat{j} + 7\hat{k}$
- $\vec{OB} = \mathbf{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$
- $\vec{OC} = \mathbf{c} = 3\hat{i} + 10\hat{j} - 1\hat{k}$
Step 2: Calculating Directional Vectors
Collinearity implies that the vectors $\vec{AB}$ and $\vec{BC}$ are parallel (scalar multiples of each other). We calculate these vectors by subtracting the initial point from the terminal point:
$\vec{AB} = \vec{OB} - \vec{OA}$
$\vec{AB} = (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k} = 1\hat{i} + 4\hat{j} - 4\hat{k}$
$\vec{BC} = \vec{OC} - \vec{OB}$
$\vec{BC} = (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k} = 1\hat{i} + 4\hat{j} - 4\hat{k}$
[By the Triangle Law of Vector Addition.]Step 3: Determining Collinearity via Scalar Multiplicity
For points to be collinear, the vector connecting two segments must be a scalar multiple of the other. Here, we observe:
$\vec{AB} = \lambda \vec{BC}$
$1\hat{i} + 4\hat{j} - 4\hat{k} = \lambda(1\hat{i} + 4\hat{j} - 4\hat{k})$
Equating components:
- $1 = \lambda(1) \implies \lambda = 1$
- $4 = \lambda(4) \implies \lambda = 1$
- $-4 = \lambda(-4) \implies \lambda = 1$
Since $\lambda = 1$ satisfies all components, $\vec{AB} = \vec{BC}$.
Step 4: Geometrical Representation
Conclusion
Because $\vec{AB}$ and $\vec{BC}$ share a common point $B$ and possess the same direction vector (they are parallel), the points $A$, $B$, and $C$ necessarily lie on the same straight line.
Final Solution: Since $\vec{AB} = \vec{BC}$ and the segments share a common point $B$, the points $A$, $B$, and $C$ are collinear.
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 $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.
- 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.
- 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 $\vec{BC}$].
- 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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