Question

The position vectors of the vertices of a triangle are \( 4\hat{i} + 5\hat{j} + \hat{k} \), \( 2\hat{i} + 4\hat{j} - \hat{k} \) and \( 3\hat{i} + 6\hat{j} - 3\hat{k} \). Then the triangle is

• A. Right-angled but not isosceles
• B. Isosceles but not right-angled
• C. Right-angled isosceles
• D. Equilateral

Hint:

Use the dot product to check for right angles between vectors. If the dot product is zero, the vectors are perpendicular, and the triangle is right-angled.

Answer 1

The Correct Option is C

Step 1: Treating \( A(4,5,1) \), \( B(2,4,-1) \), \( C(3,6,-3) \) as points, the distance formula gives \( AB=\sqrt{(4-2)^2+(5-4)^2+(1-(-1))^2}=\sqrt{4+1+4}=\sqrt9=3 \).
Step 2: Similarly, \( BC=\sqrt{(2-3)^2+(4-6)^2+(-1-(-3))^2}=\sqrt{1+4+4}=\sqrt9=3 \), and \( AC=\sqrt{(4-3)^2+(5-6)^2+(1-(-3))^2}=\sqrt{1+1+16}=\sqrt{18}=3\sqrt2 \).
Step 3: Since \( AB=BC=3 \), the triangle is isosceles, and since \( AB^2+BC^2=9+9=18=AC^2 \), the Pythagorean relation holds, confirming a right angle at \( B \).
\[ \boxed{\text{Right-angled isosceles triangle}} \]

Answer 2

A third way to classify the triangle is to compute the cosine of each angle directly from the dot product of the vectors meeting at that vertex, which confirms both the right angle and the equal sides independently.

At vertex \( B \), the vectors to the other two points are \( \overrightarrow{BA}=(2,1,2) \) and \( \overrightarrow{BC}=(1,2,-2) \). Their dot product is \( (2)(1)+(1)(2)+(2)(-2)=2+2-4=0 \), so \( \cos B=0 \), meaning \( B=90^\circ \). At vertex \( A \), the vectors are \( \overrightarrow{AB}=(-2,-1,-2) \) and \( \overrightarrow{AC}=(-1,1,-4) \), with \( |\overrightarrow{AB}|=3 \) and \( |\overrightarrow{AC}|=3\sqrt2 \), and dot product \( (-2)(-1)+(-1)(1)+(-2)(-4)=2-1+8=9 \), giving \( \cos A=\dfrac{9}{3\cdot3\sqrt2}=\dfrac{9}{9\sqrt2}=\dfrac{1}{\sqrt2} \), so \( A=45^\circ \). Since the angles of a triangle sum to \( 180^\circ \), \( C=180^\circ-90^\circ-45^\circ=45^\circ \).

1. Right-angled but not isosceles: The angle at \( B \) is indeed \( 90^\circ \), but with \( A=C=45^\circ \), the sides opposite these equal angles (\( BC \) and \( AB \)) must also be equal, so it is isosceles too, ruling this option out.
2. Isosceles but not right-angled: The two equal base angles of \( 45^\circ \) do make it isosceles, but the third angle is exactly \( 90^\circ \), so it is right-angled as well, ruling this out.
3. Right-angled isosceles: With angles \( 90^\circ,45^\circ,45^\circ \), this is precisely a right-angled isosceles triangle, matching both properties at once.
4. Equilateral: An equilateral triangle would need all angles equal to \( 60^\circ \), but here the angles are \( 90^\circ,45^\circ,45^\circ \), so it is not equilateral.

Computing all three angles directly confirms the classic \( 90^\circ, 45^\circ, 45^\circ \) angle pattern of a right-angled isosceles triangle.

Therefore, the correct answer is Right-angled isosceles.