Question

If the area of the triangle formed by the points \((0,0,0)\), \((1,1,1)\) and \((t,2t,3t)\) is \(\sqrt6\), then the sum of squares of all possible values of \(t\) is:

• A. 13
• B. 5
• C. 20
• D. 8

Hint:

Area of a triangle in 3D is half the magnitude of the cross product of two side vectors.

Answer 1

The Correct Option is D

Concept: Area of triangle formed by vectors \(\vec a\) and \(\vec b\): \[ A=\frac12 |\vec a\times \vec b|. \]

Step 1: Construct vectors.
\[ \vec a=(1,1,1), \] \[ \vec b=(t,2t,3t). \]

Step 2: Find cross product.
\[ \vec a\times \vec b = (t,-2t,t). \] Magnitude \[ = \sqrt{t^2+4t^2+t^2} = |t|\sqrt6. \]

Step 3: Use area condition.
\[ \frac12|t|\sqrt6 = \sqrt6. \] \[ |t|=2. \] Thus \[ t=\pm2. \]

Step 4: Compute sum of squares.
\[ 2^2+(-2)^2 = 8. \] Hence \[ \boxed{8}. \]