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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.
Solution :
Step 1: Assigning Position Vectors
Let the vertices of the triangle be $A$, $B$, and $C$. We represent these points by their position vectors with respect to the origin:
- $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$
- $\vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}$
- $\vec{c} = 3\hat{i} - 4\hat{j} - 4\hat{k}$
Step 2: Calculating Side Vectors
To determine the sides of the triangle, we calculate the displacement vectors $\vec{AB}$, $\vec{BC}$, and $\vec{CA}$ [By the vector definition of displacement, $\vec{PQ} = \vec{pos}(Q) - \vec{pos}(P)$]:
$\vec{AB} = \vec{b} - \vec{a} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{CA} = \vec{a} - \vec{c} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$
Step 3: Calculating Magnitudes (Squared)
We calculate the squared magnitude of each side to apply the Converse of Pythagoras Theorem [Magnitude $\|\vec{r}\| = \sqrt{x^2 + y^2 + z^2}$]:
| Side | Squared Magnitude |
|---|---|
| $|\vec{AB}|^2$ | $(-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41$ |
| $|\vec{BC}|^2$ | $(2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6$ |
| $|\vec{CA}|^2$ | $(-1)^2 + (3)^2 + (5)^2 = 1 + 9 + 25 = 35$ |
Step 4: Verification of Right-Angled Triangle Criterion
For a triangle to be right-angled, the square of the longest side must equal the sum of the squares of the other two sides [Pythagorean Theorem].
Comparing the values: $41 = 6 + 35$.
Since $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$, the triangle satisfies the Pythagorean condition.
Visual Representation of the Geometry
Final Solution: Since $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$, the vectors form the vertices of a right-angled triangle, specifically right-angled at vertex $C$.
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 $\vec{BC}$].
- Q16: Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.
- Q18: If $\vec{a}$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is unit vector if
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