JEE Advanced 2016 · Paper 2 · Mathematics Q50

← JEE Advanced 2016 · Paper 2

MathematicsMultiple correct+4 / -2Question 50 of 54

Let $\hat u=u_1\hat i+u_2\hat j+u_3\hat k$ be a unit vector in $\mathbb R^3$ and $\hat w=\dfrac1{\sqrt6}(\hat i+\hat j+2\hat k)$ . Given that there exists a vector $\vec v$ in $\mathbb R^3$ such that $|\hat u\times\vec v|=1$ and $\hat w\cdot(\hat u\times\vec v)=1$ . Which of the following statement(s) is(are) correct?

There is exactly one choice for such

\vec v

There are infinitely many choices for such

\vec v

\hat u

lies in the

xy

-plane then

|u_1|=|u_2|

\hat u

lies in the

xz

-plane then

2|u_1|=|u_3|

Answer and worked solution

Official answer: B, C

Combine the conditions. $|\hat u\times\vec v|=1$ and $\hat w\cdot(\hat u\times\vec v)=1$ with $\hat w$ a unit vector force
$\hat u\times\vec v=\hat w,$
which needs $\hat u\perp\hat w$ .

(A), (B) Any $\vec v+\lambda\hat u$ works too, so there are infinitely many choices. (B) is true, (A) is false.
(C) $\hat u$ in the $xy$ -plane: $u_1+u_2=0$ , so $|u_1|=|u_2|$ . True.
(D) $\hat u$ in the $xz$ -plane: $u_1+2u_3=0$ , so $|u_1|=2|u_3|$ , not the reverse. False.

(B), (C)

Topics:Cross product Dot product

← Previous question Next question →

JEE Advanced 2016 · Paper 2, Mathematics question 50