Question

Are the three points whose position vectors are \(2\hat{i} + 3\hat{j} - 4\hat{k}\), \(\hat{i} - 2\hat{j} + 3\hat{k}\) and \(-7\hat{j} + 10\hat{k}\) collinear?

• A. Yes
• B. No
• C. Cannot be determined
• D. None of these

Hint:

Form two vectors from a common point and check whether one is a scalar multiple of the other.

Answer 1

The Correct Option is A

Step 1: Three points are collinear if the vector from the first to the second is a scalar multiple of the vector from the first to the third. Label the points \[A = 2\hat{i} + 3\hat{j} - 4\hat{k},\quad B = \hat{i} - 2\hat{j} + 3\hat{k},\quad C = -7\hat{j} + 10\hat{k}.\]

Step 2: Compute \(\vec{AB} = B - A\): \[\vec{AB} = (1-2)\hat{i} + (-2-3)\hat{j} + (3-(-4))\hat{k} = -\hat{i} - 5\hat{j} + 7\hat{k}.\]

Step 3: Compute \(\vec{AC} = C - A\): \[\vec{AC} = (0-2)\hat{i} + (-7-3)\hat{j} + (10-(-4))\hat{k} = -2\hat{i} - 10\hat{j} + 14\hat{k}.\]

Step 4: Compare the two vectors: \[\vec{AC} = -2\hat{i} - 10\hat{j} + 14\hat{k} = 2\,(-\hat{i} - 5\hat{j} + 7\hat{k}) = 2\,\vec{AB}.\] Since \(\vec{AC}\) is exactly twice \(\vec{AB}\), the two vectors are parallel and share the common point \(A\).

Step 5: Parallel vectors from a common point lie along the same line, so the three points are collinear. The answer is (A), Yes.

\[\boxed{\text{Yes, the points are collinear}}\]