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JEE Mathematics 2023 Question with Solution

Let O be the origin and the position vector of the point P be −i^−2j^+3k^-\hat{i}-2\hat{j}+3\hat{k}. If the position vectors of A, B, and C are −2i^+j^−3k^-2\hat{i}+\hat{j}-3\hat{k}, 2i^+4j^−2k^2\hat{i}+4\hat{j}-2\hat{k}, and −4i^+2j^−k^-4\hat{i}+2\hat{j}-\hat{k} respectively, then the projection of vector OP⃗\vec{OP} on a vector perpendicular to vectors AB⃗\vec{AB} and AC⃗\vec{AC} is:

  • A

    103\frac{10}{3}

  • B

    83\frac{8}{3}

  • C

    73\frac{7}{3}

  • D

    33

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: P(−1,−2,3)P(-1,-2,3), A(−2,1,−3)A(-2,1,-3), B(2,4,−2)B(2,4,-2), and C(−4,2,−1)C(-4,2,-1).

Find: The projection of OP⃗\vec{OP} on a vector perpendicular to AB⃗\vec{AB} and AC⃗\vec{AC}.

A vector perpendicular to both AB⃗\vec{AB} and AC⃗\vec{AC} is along AB⃗×AC⃗\vec{AB} \times \vec{AC}.

First compute

AB⃗=⟨4,3,1⟩,AC⃗=⟨−2,1,2⟩\vec{AB}=\langle 4,3,1 \rangle, \qquad \vec{AC}=\langle -2,1,2 \rangle

Then

AB⃗×AC⃗=∣i^j^k^431−212∣\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix}

so

AB⃗×AC⃗=i^(3⋅2−1⋅1)−j^(4⋅2−1⋅−2)+k^(4⋅1−3⋅−2)\vec{AB} \times \vec{AC} = \hat{i}(3 \cdot 2 - 1 \cdot 1) - \hat{j}(4 \cdot 2 - 1 \cdot -2) + \hat{k}(4 \cdot 1 - 3 \cdot -2) AB⃗×AC⃗=5i^−10j^+10k^\vec{AB} \times \vec{AC} = 5\hat{i} - 10\hat{j} + 10\hat{k}

Its magnitude is

∣AB⃗×AC⃗∣=52+(−10)2+102=225=15\left|\vec{AB} \times \vec{AC}\right| = \sqrt{5^2 + (-10)^2 + 10^2} = \sqrt{225} = 15

Hence a unit vector perpendicular to both is

AB⃗×AC⃗∣AB⃗×AC⃗∣=5i^−10j^+10k^15\frac{\vec{AB} \times \vec{AC}}{\left|\vec{AB} \times \vec{AC}\right|} = \frac{5\hat{i} - 10\hat{j} + 10\hat{k}}{15}

Now

OP⃗=−i^−2j^+3k^\vec{OP} = -\hat{i} - 2\hat{j} + 3\hat{k}

Therefore, the required projection is

OP⃗⋅AB⃗×AC⃗∣AB⃗×AC⃗∣\vec{OP} \cdot \frac{\vec{AB} \times \vec{AC}}{\left|\vec{AB} \times \vec{AC}\right|} =(−1,−2,3)⋅(515,−1015,1015)= (-1,-2,3) \cdot \left(\frac{5}{15},\frac{-10}{15},\frac{10}{15}\right) =−5+20+3015=4515=3= \frac{-5 + 20 + 30}{15} = \frac{45}{15} = 3

The working in the solution gives the value 33, but it also states 'The Correct Option is C', which conflicts with the listed options. Since 33 corresponds to option D, the keyed label appears inconsistent. The defensible correct choice from the working is D.

Therefore, the correct option is D.

Why the cross product is used

A vector perpendicular to both AB⃗\vec{AB} and AC⃗\vec{AC} must be normal to the plane containing AA, BB, and CC. The cross product AB⃗×AC⃗\vec{AB} \times \vec{AC} gives exactly such a normal vector. To get the scalar projection of OP⃗\vec{OP} on that direction, we take the dot product of OP⃗\vec{OP} with the corresponding unit normal vector. This leads directly to the value 33, so the matching option is D.

Common mistakes

  • Using AB⃗+AC⃗\vec{AB} + \vec{AC} instead of AB⃗×AC⃗\vec{AB} \times \vec{AC}. A vector perpendicular to two given vectors is obtained from the cross product, not from addition. Always form the normal direction with AB⃗×AC⃗\vec{AB} \times \vec{AC}.

  • Computing AB⃗\vec{AB} or AC⃗\vec{AC} with reversed subtraction. Since AB⃗=OB⃗−OA⃗\vec{AB}=\vec{OB}-\vec{OA} and AC⃗=OC⃗−OA⃗\vec{AC}=\vec{OC}-\vec{OA}, sign errors here change the normal vector and the final projection. Subtract coordinates carefully in the correct order.

  • Taking the dot product with AB⃗×AC⃗\vec{AB} \times \vec{AC} directly without dividing by its magnitude. Projection on a vector direction requires a unit vector if the scalar projection is asked. First normalize by ∣AB⃗×AC⃗∣\left|\vec{AB} \times \vec{AC}\right|.

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