Let O be the origin and the position vector of the point P be - \widehat i - 2\widehat j + 3\widehat k . If the position vectors of the points A, B and C are - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k and - 4\widehat i + 2\widehat j - \widehat k respectively, then the projection of the vector \overrightarrow OP on a vector perpendicular to the vectors \overrightarrow AB and \overrightarrow AC is :

Vector Algebra PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q27 with Solution

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Correct answer: B

$\vec{OP}= -\hat i-2\hat j+3\hat k$

$\vec{OA}= -2\hat i+\hat j-3\hat k$ $\vec{OB}= 2\hat i+4\hat j-2\hat k$ $\vec{OC}= -4\hat i+2\hat j-\hat k$

We need the projection of $\vec{OP}$ on a vector perpendicular to both $\vec{AB}$ and $\vec{AC}$ .

A vector perpendicular to both $\vec{AB}$ and $\vec{AC}$ is along

$\vec{AB}\times \vec{AC}.$

$\vec{AB}=\vec{OB}-\vec{OA}=(2+2)\hat i+(4-1)\hat j+(-2+3)\hat k$ $\vec{AB}=4\hat i+3\hat j+\hat k$

$\vec{AC}=\vec{OC}-\vec{OA}=(-4+2)\hat i+(2-1)\hat j+(-1+3)\hat k$ $\vec{AC}=-2\hat i+\hat j+2\hat k$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & 3 & 1\\ -2 & 1 & 2 \end{vmatrix}$$ $$=\hat i(3\cdot 2-1\cdot 1)-\hat j(4\cdot 2-1\cdot(-2))+\hat k(4\cdot 1-3\cdot(-2))$$ $$=5\hat i-10\hat j+10\hat k$$ So a perpendicular vector is $$\vec{n}=5\hat i-10\hat j+10\hat k=5(\hat i-2\hat j+2\hat k).$$ Hence we can take the simpler direction vector $$\vec{n}=\hat i-2\hat j+2\hat k.$$ --- 4. **Find the scalar projection of $\vec{OP}$ on this direction** Scalar projection of $\vec{OP}$ on $\vec{n}$ is $$\frac{\vec{OP}\cdot \vec{n}}{|\vec{n}|}.$$ Now, $$\vec{OP}\cdot \vec{n}=(-1)(1)+(-2)(-2)+(3)(2)$$ $$=-1+4+6=9$$ and $$|\vec{n}|=\sqrt{1^2+(-2)^2+2^2}=\sqrt{1+4+4}=3.$$ Therefore projection is $$\frac{9}{3}=3.$$ --- 5. **Check options** - A: $\frac{7}{3}$ ❌ - B: $3$ ✅ - C: $\frac{10}{3}$ ❌ - D: $\frac{8}{3}$ ❌ So the correct answer is **Option B**.