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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.
Solution :
Step 1: Determining the Side Vectors
To determine if the points $A$, $B$, and $C$ form a right-angled triangle, we must calculate the vectors representing the sides of the triangle: $\vec{AB}$, $\vec{BC}$, and $\vec{CA}$.
Given position vectors:
- $\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}$
- $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$
- $\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}$
Calculating the displacement vectors [By Vector Subtraction Rule]:
- $\vec{AB} = \vec{b} - \vec{a} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$
- $\vec{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
- $\vec{CA} = \vec{a} - \vec{c} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$
Step 2: Calculating Magnitudes of the Sides
The squared magnitudes of the vectors represent the squares of the side lengths $|\vec{AB}|^2, |\vec{BC}|^2, |\vec{CA}|^2$ [Per the Definition of Magnitude of a Vector]:
| Vector | Magnitude Squared ($|\vec{v}|^2$) |
|---|---|
| $\vec{AB}$ | $(-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35$ |
| $\vec{BC}$ | $(-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41$ |
| $\vec{CA}$ | $2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$ |
Step 3: Verifying the Pythagorean Theorem
According to the Converse of the Pythagorean Theorem, if the sum of the squares of two sides equals the square of the third side, the triangle is right-angled.
Summing the two smaller magnitudes: $|\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41$
Comparing with the largest magnitude: $|\vec{BC}|^2 = 41$
Since $|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$, the triangle satisfies the condition for a right-angled triangle, specifically with the right angle at vertex $A$ (as $\vec{AB} \cdot \vec{CA} = -2 - 3 + 5 = 0$).
Final Solution: Since the magnitudes satisfy the relation $|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$, the points A, B, and C form a right-angled triangle with the right angle at vertex A.
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.
- 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.
- 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).
- 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
Chapters in CBSE - Class 12 Mathematics
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