Let {u} = {u₁ i} + {u₂ j} + {u₃ k} be a unit vector in R² and {w} = {1}{{6}} ( {i} + {j} + {2k} ).Given n that there exists a vector {v} in R³ such that |{u} × {v}| = 1 and {w} · ( {u} × {v} ) = 1.which of the following statement(s) is(are) correct? [JEE Adv 2016 P2]

There is exactly one choice for such {v}

There are infinitely many choice for such {v}

If {u} lies in the xy-plane then |u₁| = |u₂|

If {u} lies in the xy-plane then 2|u₁| = |u₃|

Test Index

Vector Algebra

Section: Mathematics

Question : 17 of 38

Marks: +1, -0

Let

\hat{u} = \hat{u_1 i} + \hat{u_2 j} + \hat{u_3 k}

R^2

\hat{w} = \frac{1}{\sqrt{6}} ( \hat{i} + \hat{j} + \hat{2k} )

\vec{v}

R^3

|\hat{u} \times \vec{v}| = 1

\hat{w} \cdot ( \hat{u} \times \vec{v} ) = 1

Solution:

|\hat{w}| |\hat{u} \times \hat{v}| \cos \phi = 1 \Rightarrow \phi = 0

\Rightarrow \hat{u} \times \hat{v} = \hat{w}

also

|\vec{v}| \sin \theta = 1

⇒ there may be infinite vectors

\vec{v} \vec{OP}

such that P is always 1 unit dist. from

\hat{u}

For option (C):

\hat{u} \times \hat{v} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & 0 \\ v_1 & v_2 & v_3 \end{vmatrix}

\hat{w} = (u_2 v_3) \hat{i} - (u_1 v_3) \hat{j} + (u_1 v_2 - u_2 v_1) \hat{k}

u_2 v_3 = \frac{1}{\sqrt{6}}, -u_1 v_3 = \frac{1}{\sqrt{6}} \Rightarrow |u_1| = |u_2|

For option (D):

\hat{u} \times \hat{v} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & 0 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

\hat{w} = (-v_2 u_3) \hat{i} - (u_1 v_3 - u_3 v_1) \hat{j} + (u_1 v_2) \hat{k}

-v_2 u_3 = \frac{1}{\sqrt{6}}, u_1 v_2 = \frac{2}{\sqrt{6}}

\Rightarrow 2 |u_3| = |u_1|

So (D) is wrong

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