Let u^=u1i^+u2j^+u3k^ be a unit vector in R3 and w^=16 i^+ j^+2k^. Given 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 ?

(C) If u^ lies in the xy - plane then u1=u2

(A) There is exactly one choice for such v→

(B) There are infinitely many choice for such v→

(C) If u^ lies in the xy - plane then u1=u2

(D) If u^ lies in the xz - plane then 2u1=u3

JEE Advanced · Mathematics · 30. Vector Algebra

Let $\hat{u} = u_{1} \hat{i} + u_{2} \hat{j} + u_{3} \hat{k}$ be a unit vector in $R^{3}$ and $\hat{w} = \frac{1}{\sqrt{6}} (\hat{i} + \hat{j} + 2 \hat{k}) .$ Given that there exists a vector $\vec{v}$ in $R^{3}$ such that $|\hat{u} \times \vec{v}| = 1$ and $\hat{w} . (\hat{u} \times \vec{v}) = 1 .$ Which of the following statement(s) is (are) correct ?

A There is exactly one choice for such $\vec{v}$
B There are infinitely many choice for such $\vec{v}$
C If $\hat{u}$ lies in the xy - plane then $|u_{1}| = |u_{2}|$
D If $\hat{u}$ lies in the xz - plane then $2 |u_{1}| = |u_{3}|$

Verified Solution

Answer & Solution

Correct Answer

(C) If $\hat{u}$ lies in the xy - plane then $|u_{1}| = |u_{2}|$

Step-by-step Solution

Detailed explanation

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

Let

ϕ

is angle between

\hat{w} & (\hat{u} \times \vec{v})

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

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

\Rightarrow

There may be infinite vectors

\vec{v}

Satisfying this condition
If

\hat{u}

his in

x y

plane :

\hat{u} \times \vec{v} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ u_{1} & u_{2} & 0 \\ v_{1} & v_{2} & v_{3} \end{matrix}|

= \hat{w}

\Rightarrow u_{3} = 0

Let

\vec{v} = v_{1} \hat{i} + v_{2} \hat{j} + v_{3} \hat{k}

\hat{w} = (u_{2} v_{3}) \hat{i} - (u_{1} v_{3}) \hat{j} + (u_{1} v_{2} - u_{2} v_{1}) \hat{k}

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

u_{2} v_{3} = \frac{1}{\sqrt{6}}, - u_{1} v_{3} = \frac{1}{\sqrt{6}} \Rightarrow |u_{1}| = |u_{2}|

\hat{u}

his in

x z

plane

\Rightarrow u_{2} = 0

\hat{u} \times \vec{v} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ u_{1} & 0 & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix}|

\hat{w} = (- v_{2} u_{3}) \hat{i} - (u_{1} v_{3} - u_{3} v_{1}) \hat{j} + (u_{1} v_{2}) \hat{k}

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

- v_{2} u_{3} = \frac{1}{\sqrt{6}}, u_{1} v_{2} = \frac{2}{\sqrt{6}}

\Rightarrow

2 |u_{3}| = |u_{1}|

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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