PrepRung
Sign inSign up free

JEE Advanced 2024 Paper 1 · Mathematics · Vector Algebra · medium

Let vec a=hat i+2hat j+2hat k, vec b=2hat i+hat j-2hat k, vec c=2hat i-hat j+3hat k. Match the entries of List-I with List-II. List-I P. |vec a×vec…

Single correct option

Let a⃗=i^+2j^+2k^\vec a=\hat i+2\hat j+2\hat k, b⃗=2i^+j^−2k^\vec b=2\hat i+\hat j-2\hat k, c⃗=2i^−j^+3k^\vec c=2\hat i-\hat j+3\hat k. Match the entries of List-I with List-II.

List-I

P. ∣a⃗×b⃗∣|\vec a\times\vec b|

Q. Volume of the tetrahedron with edges OA→=a⃗\overrightarrow{OA}=\vec a, OB→=b⃗\overrightarrow{OB}=\vec b, OC→=c⃗\overrightarrow{OC}=\vec c meeting at the origin OO

R. Scalar projection of c⃗\vec c on a⃗\vec a

S. Distance between the parallel planes 2x−2y+z=32x-2y+z=3 and 4x−4y+2z=−94x-4y+2z=-9

List-II

(1) 52\dfrac52

(2) 92\dfrac92

(3) 99

(4) 22

(5) 33
  1. A. P→3, Q→2, R→1, S→4
  2. B. P→3, Q→2, R→4, S→1
  3. C. P→2, Q→3, R→4, S→1
  4. D. P→3, Q→5, R→4, S→1
Show answer and solution

Answer: (B) P→3, Q→2, R→4, S→1

P: a⃗⋅b⃗=2+2−4=0\vec a\cdot\vec b=2+2-4=0 and ∣a⃗∣=∣b⃗∣=3|\vec a|=|\vec b|=3, so ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡90∘=9|\vec a\times\vec b|=|\vec a||\vec b|\sin90^\circ=9, i.e. (3). Q: a⃗×b⃗=(−6, 6, −3)\vec a\times\vec b=(-6,\,6,\,-3) and c⃗⋅(a⃗×b⃗)=−12−6−9=−27\vec c\cdot(\vec a\times\vec b)=-12-6-9=-27, so [a⃗ b⃗ c⃗]=−27[\vec a\ \vec b\ \vec c]=-27 and the tetrahedron volume is 276=92\tfrac{27}{6}=\tfrac92, i.e. (2). R: c⃗⋅a⃗=2−2+6=6\vec c\cdot\vec a=2-2+6=6, and dividing by ∣a⃗∣=3|\vec a|=3 gives 22, i.e. (4). S: rewrite the second plane as 2x−2y+z=−922x-2y+z=-\tfrac92; the distance is ∣3−(−9/2)∣4+4+1=15/23=52\dfrac{|3-(-9/2)|}{\sqrt{4+4+1}}=\dfrac{15/2}{3}=\dfrac52, i.e. (1). So P→3, Q→2, R→4, S→1.
Report a mistake

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake