Question

The vertices of triangle \( \Delta ABC \) are \( A(2, 3, k) \), \( B(-1, k, -1) \), and \( C(4, -3, 2) \). If \( AB = AC \) and \( k>0 \), then the triangle \( ABC \) is:

• A. an equilateral triangle
• B. a right-angled isosceles triangle
• C. an isosceles triangle but not right angled
• D. an obtuse angled isosceles triangle \bigskip

Hint:

When given points in 3D space, use the distance formula to find the lengths of the sides of the triangle and then apply the dot product to check for right angles.

Answer 1

The Correct Option is B

We are given the points \( A(2, 3, k) \), \( B(-1, k, -1) \), and \( C(4, -3, 2) \), and we need to prove that \( AB = AC \) to show the type of triangle. Step 1: Find the distance \( AB \) The distance between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) in 3-dimensional space is given by the formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] The distance between points \( A(2, 3, k) \) and \( B(-1, k, -1) \) is: \[ AB = \sqrt{((-1) - 2)^2 + (k - 3)^2 + (-1 - k)^2} \] \[ AB = \sqrt{(-3)^2 + (k - 3)^2 + (-1 - k)^2} \] \[ AB = \sqrt{9 + (k - 3)^2 + (k + 1)^2} \] Expanding the squares: \[ AB = \sqrt{9 + (k^2 - 6k + 9) + (k^2 + 2k + 1)} \] \[ AB = \sqrt{9 + 2k^2 - 4k + 10} \] \[ AB = \sqrt{2k^2 - 4k + 19} \] Step 2: Find the distance \( AC \) Similarly, the distance between points \( A(2, 3, k) \) and \( C(4, -3, 2) \) is: \[ AC = \sqrt{(4 - 2)^2 + (-3 - 3)^2 + (2 - k)^2} \] \[ AC = \sqrt{2^2 + (-6)^2 + (2 - k)^2} \] \[ AC = \sqrt{4 + 36 + (2 - k)^2} \] Expanding the square: \[ AC = \sqrt{4 + 36 + (k^2 - 4k + 4)} \] \[ AC = \sqrt{k^2 - 4k + 44} \] Step 3: Set \( AB = AC \) Since \( AB = AC \), we equate the two distances: \[ \sqrt{2k^2 - 4k + 19} = \sqrt{k^2 - 4k + 44} \] Squaring both sides: \[ 2k^2 - 4k + 19 = k^2 - 4k + 44 \] Simplifying: \[ 2k^2 - k^2 = 44 - 19 \] \[ k^2 = 25 \] \[ k = 5 \] Step 4: Find the angle Now, we will use the fact that the triangle is isosceles (since \( AB = AC \)) to find whether it is a right-angled triangle. The condition for a right-angled triangle is that the dot product of two vectors representing two sides of the triangle is zero. The vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are: \[ \overrightarrow{AB} = B - A = (-1, 5, -1) - (2, 3, 5) = (-3, 2, -6) \] \[ \overrightarrow{AC} = C - A = (4, -3, 2) - (2, 3, 5) = (2, -6, -3) \] The dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) is: \[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (-3)(2) + (2)(-6) + (-6)(-3) \] \[ \overrightarrow{AB} \cdot \overrightarrow{AC} = -6 - 12 + 18 = 0 \] Since the dot product is zero, the angle between \( AB \) and \( AC \) is \( 90^\circ \), confirming that the triangle is a right-angled isosceles triangle. Thus, the triangle is a right-angled isosceles triangle. \bigskip