Show that the points $A(1,\, 2,\, 7)$, $B(2,\, 6,\, 3)$ and $C(3,\, 1…

Question 26 of 68intermediate🔧 ApplyLong Answer4 marks

Show that the points $A(1,\, 2,\, 7)$ , $B(2,\, 6,\, 3)$ and $C(3,\, 10,\, -1)$ are collinear.

Correct Answer

The points $A(1,\, 2,\, 7)$ , $B(2,\, 6,\, 3)$ and $C(3,\, 10,\, -1)$ are collinear.

Exercise: EXERCISE 10.3 | Q: 16 | (Chapter: 25)

For More Understanding

Explanation

This question tests the concept of collinearity using vector algebra. From the textbook context, we learn that points $A$ , $B$ , $C$ are collinear if $|AC| = |AB| + |BC|$ . This is derived from the triangle inequality where equality holds only when points lie on a straight line.

The solution involves finding position vectors, calculating displacement vectors $\vec{AB}$ , $\vec{BC}$ , and $\vec{AC}$ , then computing their magnitudes to verify the relationship.

Solution Steps

Step 1: Find the vectors $\vec{AB}$ , $\vec{BC}$ and $\vec{AC}$ .

$\vec{AB} = (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k} = \hat{i} + 4\hat{j} - 4\hat{k}$

$\vec{BC} = (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k} = \hat{i} + 4\hat{j} - 4\hat{k}$

$\vec{AC} = (3-1)\hat{i} + (10-2)\hat{j} + (-1-7)\hat{k} = 2\hat{i} + 8\hat{j} - 8\hat{k}$
Step 2: Calculate the magnitudes.

$|\vec{AB}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33}$

$|\vec{BC}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33}$

$|\vec{AC}| = \sqrt{2^2 + 8^2 + (-8)^2} = \sqrt{4 + 64 + 64} = \sqrt{132} = 2\sqrt{33}$
Step 3: Verify the relationship between magnitudes.

$|\vec{AB}| + |\vec{BC}| = \sqrt{33} + \sqrt{33} = 2\sqrt{33} = |\vec{AC}|$

Therefore, $|\vec{AC}| = |\vec{AB}| + |\vec{BC}|$
Step 4: Conclusion.

Since $|\vec{AC}| = |\vec{AB}| + |\vec{BC}|$ , the points $A$ , $B$ and $C$ are collinear.

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