Let P be a point in the plane of the vectors AB = 3i + j - k and AC = i - j + 3k such that P is equidistant from the lines AB and AC. If |AP| = (√(5))/2, then the area of the triangle ABP is:

Vector Algebra: JEE Main 2026 Mathematics question

JEE Main 2026 · 28 Jan, Evening shift · Q19

JEE Main 202628 Jan, Evening shiftMathematicsMediumVector Algebra184× since 2002

Let P be a point in the plane of the vectors

\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}

and

\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}

such that P is equidistant from the lines AB and AC. If

|\vec{AP}| = \frac{\sqrt{5}}{2}

, then the area of the triangle ABP is :

JEE Main 2026 · 28 Jan, Evening shift · Q19

JEE Main 202628 Jan, Evening shiftMathematicsMediumVector Algebra184× since 2002

Let P be a point in the plane of the vectors

\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}

and

\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}

such that P is equidistant from the lines AB and AC. If

|\vec{AP}| = \frac{\sqrt{5}}{2}

, then the area of the triangle ABP is :