Question:
If the position vectors of three points A, B and C are \( 3\hat{i} + \hat{j} \), \( 5\hat{i} + 6\hat{j} - 3\hat{k} \) and \( 4\hat{j} \) respectively, then show that they form an isosceles triangle.
If the position vectors of three points A, B and C are \( 3\hat{i} + \hat{j} \), \( 5\hat{i} + 6\hat{j} - 3\hat{k} \) and \( 4\hat{j} \) respectively, then show that they form an isosceles triangle.
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Always simplify magnitudes under the square root, but don't feel pressured to convert to decimals; radical form is preferred.
The vector \( \vec{AC} \) can also be found using the triangle law: \( \vec{AC} = \vec{AB} + \vec{BC} \).
The vector \( \vec{AC} \) can also be found using the triangle law: \( \vec{AC} = \vec{AB} + \vec{BC} \).
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• A triangle is isosceles if any two of its sides are equal in length.
• The length of a side is the magnitude of the vector representing that side.
• Vector \( \vec{AB} = \text{P.V. of B} - \text{P.V. of A} \).
• Magnitude of \( \vec{v} = x\hat{i} + y\hat{j} + z\hat{k} \) is \( |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \).
Step 1: Find the vector representation for each side
Given: \( \vec{A} = 3\hat{i} + \hat{j} \), \( \vec{B} = 5\hat{i} + 6\hat{j} - 3\hat{k} \), \( \vec{C} = 4\hat{j} \).
\[ \vec{AB} = (5-3)\hat{i} + (6-1)\hat{j} + (-3-0)\hat{k} = 2\hat{i} + 5\hat{j} - 3\hat{k} \] \[ \vec{BC} = (0-5)\hat{i} + (4-6)\hat{j} + (0 - (-3))\hat{k} = -5\hat{i} - 2\hat{j} + 3\hat{k} \] \[ \vec{AC} = (0-3)\hat{i} + (4-1)\hat{j} + (0-0)\hat{k} = -3\hat{i} + 3\hat{j} + 0\hat{k} \]
Step 2: Calculate the length (magnitude) of each side
Length of \( AB = |\vec{AB}| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \) units.
Length of \( BC = |\vec{BC}| = \sqrt{(-5)^2 + (-2)^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38} \) units.
Length of \( AC = |\vec{AC}| = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9} = \sqrt{18} \) units.
Step 3: Draw a conclusion
Comparing the lengths calculated in Step 2, we see that:
\( AB = BC = \sqrt{38} \) units.
Since two sides of the triangle are equal in length, \( \Delta ABC \) is an isosceles triangle.
• A triangle is isosceles if any two of its sides are equal in length.
• The length of a side is the magnitude of the vector representing that side.
• Vector \( \vec{AB} = \text{P.V. of B} - \text{P.V. of A} \).
• Magnitude of \( \vec{v} = x\hat{i} + y\hat{j} + z\hat{k} \) is \( |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \).
Step 1: Find the vector representation for each side
Given: \( \vec{A} = 3\hat{i} + \hat{j} \), \( \vec{B} = 5\hat{i} + 6\hat{j} - 3\hat{k} \), \( \vec{C} = 4\hat{j} \).
\[ \vec{AB} = (5-3)\hat{i} + (6-1)\hat{j} + (-3-0)\hat{k} = 2\hat{i} + 5\hat{j} - 3\hat{k} \] \[ \vec{BC} = (0-5)\hat{i} + (4-6)\hat{j} + (0 - (-3))\hat{k} = -5\hat{i} - 2\hat{j} + 3\hat{k} \] \[ \vec{AC} = (0-3)\hat{i} + (4-1)\hat{j} + (0-0)\hat{k} = -3\hat{i} + 3\hat{j} + 0\hat{k} \]
Step 2: Calculate the length (magnitude) of each side
Length of \( AB = |\vec{AB}| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \) units.
Length of \( BC = |\vec{BC}| = \sqrt{(-5)^2 + (-2)^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38} \) units.
Length of \( AC = |\vec{AC}| = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9} = \sqrt{18} \) units.
Step 3: Draw a conclusion
Comparing the lengths calculated in Step 2, we see that:
\( AB = BC = \sqrt{38} \) units.
Since two sides of the triangle are equal in length, \( \Delta ABC \) is an isosceles triangle.
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