PrepRung
Sign inSign up free

JEE Advanced 2025 Paper 2 · Mathematics · Vector Algebra · medium

Let vec a=hat i+2hat j-hat k, vec b=2hat i-hat j+hat k and vec c=hat i+2hat j+3hat k. A vector vec r satisfies vec r×vec a=vec b×vec a and vec r·vec…

Numerical answer

Let a⃗=i^+2j^−k^\vec a=\hat i+2\hat j-\hat k, b⃗=2i^−j^+k^\vec b=2\hat i-\hat j+\hat k and c⃗=i^+2j^+3k^\vec c=\hat i+2\hat j+3\hat k. A vector r⃗\vec r satisfies r⃗×a⃗=b⃗×a⃗\vec r\times\vec a=\vec b\times\vec a and r⃗⋅c⃗=0\vec r\cdot\vec c=0. Then ∣r⃗∣2=|\vec r|^2= ______. Give the answer correct to two decimal places.
Show answer and solution

Answer: 22.50

From (r⃗−b⃗)×a⃗=0⃗(\vec r-\vec b)\times\vec a=\vec 0, r⃗−b⃗\vec r-\vec b is parallel to a⃗\vec a: r⃗=b⃗+λa⃗\vec r=\vec b+\lambda\vec a. Now b⃗⋅c⃗=2−2+3=3\vec b\cdot\vec c=2-2+3=3 and a⃗⋅c⃗=1+4−3=2\vec a\cdot\vec c=1+4-3=2. The condition r⃗⋅c⃗=0\vec r\cdot\vec c=0 gives 3+2λ=03+2\lambda=0, so λ=−32\lambda=-\tfrac32. Then r⃗=(2−32, −1−3, 1+32)=(12,−4,52)\vec r=(2-\tfrac32,\,-1-3,\,1+\tfrac32)=(\tfrac12,-4,\tfrac52). So ∣r⃗∣2=14+16+254=22.5|\vec r|^2=\tfrac14+16+\tfrac{25}4=22.5. A direct check confirms r⃗×a⃗=b⃗×a⃗\vec r\times\vec a=\vec b\times\vec a, so the answer is 22.5022.50.
Report a mistake

Original question written in the JEE Advanced 2025 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake