Let P be a point in the plane of the vectors AB=3 i+ j- k and AC= i- j+3 k such that P is equidistant from the lines AB and AC. If | AP|= 52, then the area of triangle ABP is:

Given: AB=3 i+ j- k, AC= i- j+3 k, and | AP|= 52. Find: The area of triangle ABP. A point equidistant from the two intersecting lines AB and AC lies on the internal angle bisector of the angle between them. Since | AB|= 9+1+1= 11, | AC|= 1+1+9= 11 the angle bisector direction is proportional to AB| AB|+ AC| AC| So we can take d= AB+ AC=4 i+0 j+2 k Its magnitude is | d|= 16+4= 20 Hence the unit vector along this direction is d= 1 20(4 i+2 k) Therefore AP= 52 d Now the area of triangle ABP is Area= 12| AB AP| From the working, | AB AP|=3 Thus Area= 12 3= 32 Therefore, the correct option is B and the area of triangle ABP is 32.

Let P be a point in the plane of the vectors AB=3 i+ j- k and AC=… — Solution | JEE Mathematics

MCQMediumJEE 2026Equation of Line in 3D

Let $P$ be a point in the plane of the vectors $\vec{AB}=3\hat{i}+\hat{j}-\hat{k} \quad \text{and} \quad \vec{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that $P$ is equidistant from the lines $AB$ and $AC$ . If $|\vec{AP}|=\frac{\sqrt5}{2}$ , then the area of triangle $ABP$ is:

A
$2$
B
$\frac{3}{2}$
C
$\frac{\sqrt{26}}{4}$
D
$\frac{\sqrt{30}}{4}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $\vec{AB}=3\hat{i}+\hat{j}-\hat{k}$ , $\vec{AC}=\hat{i}-\hat{j}+3\hat{k}$ , and $|\vec{AP}|=\frac{\sqrt5}{2}$ .

Find: The area of triangle $ABP$ .

A point equidistant from the two intersecting lines $AB$ and $AC$ lies on the internal angle bisector of the angle between them.

Since

|\vec{AB}|=\sqrt{9+1+1}=\sqrt{11}, \quad |\vec{AC}|=\sqrt{1+1+9}=\sqrt{11}

the angle bisector direction is proportional to

\frac{\vec{AB}}{|\vec{AB}|}+\frac{\vec{AC}}{|\vec{AC}|}

So we can take

\vec{d}=\vec{AB}+\vec{AC}=4\hat{i}+0\hat{j}+2\hat{k}

Its magnitude is

|\vec{d}|=\sqrt{16+4}=\sqrt{20}

Hence the unit vector along this direction is

\hat{d}=\frac{1}{\sqrt{20}}(4\hat{i}+2\hat{k})

Therefore

\vec{AP}=\frac{\sqrt5}{2}\hat{d}

Now the area of triangle $ABP$ is

\text{Area}=\frac12|\vec{AB}\times\vec{AP}|

From the working,

|\vec{AB}\times\vec{AP}|=3

Thus

\text{Area}=\frac{1}{2}\times 3=\frac{3}{2}

Therefore, the correct option is B and the area of triangle $ABP$ is $\frac{3}{2}$ .

Common mistakes

Assuming that equidistant from the lines $AB$ and $AC$ means equal distance from the points $B$ and $C$ . This is wrong because the condition is about perpendicular distance from two lines, not distance from endpoints. Use the angle bisector property for intersecting lines instead.
Using $\vec{AB}-\vec{AC}$ without checking which bisector is needed. The internal angle bisector uses the sum of the corresponding unit vectors. Here, because both given magnitudes are equal, the required direction is along $\vec{AB}+\vec{AC}$ .
Forgetting the factor $\frac12$ in the triangle area formula. The magnitude $|\vec{AB}\times\vec{AP}|$ gives the area of the parallelogram, so the area of triangle $ABP$ is half of that.

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