Let - u - u 1 - i - u 2 - j - u 3 - k be a unit vector in R 3 and - w - 1 - 6 - i - j -2- k - Given that there exists a vector v in R 3 such that - - u - v - -1 and - w - - u - v -1- Which of the following statements isare correct-

The correct options are B There are infinitely many choices for such v C If ∧ u lies in the xy plane then | u _ 1 | =| u _ 2 | ŵ.( ...

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Byju's Answer

Standard XII

Mathematics

Definition of Vector

Let ∧ u = u...

Question

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

There is exactly one choice for such

\to v

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There are infinitely many choices for such

\to v

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\land u

lies in the

x y

-plane then

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

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\land u

lies in the

x z

-plane then

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

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Solution

The correct options are
B There are infinitely many choices for such $\to v$
C If $\land u$ lies in the $x y$ -plane then $| u_{1} | = | u_{2} |$
$^w . (^u \times \to v) = 1$
$\Rightarrow | w | |^u \times \to v | c o s α = 1$
$\Rightarrow c o s α = 1$
So, $^w$ is parallel to $(^u \times \to v)$
$\Rightarrow^w ⊥^u$ and $^w ⊥ \to v$
It is given that there exists a vector $\to v,$
So, there is a vector $\to v$ for every possible $^u$
Since $^w ⊥^u$
$\Rightarrow^w .^u = 0$
$\Rightarrow u_{1} + u_{2} + 2 u_{3} = 0$
Infinitely many vectors satisfy this condition, so there are infinite choices for $^u$
$\Rightarrow$ there are infinitely many choices for $\to v$
Now,
If $^u$ lies in the $x y$ -plane $\Rightarrow u_{3} = 0 \Rightarrow | u_{1} | = | u_{2} |$
If $^u$ lies in the $x z$ -plane $\Rightarrow u_{2} = 0 \Rightarrow | u_{1} | = 2 | u_{3} |$
So, Options (B,C) are correct

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Similar questions

Q. Let

^u = u_{1}^i + u_{2}^j + u_{3}^k

be a unit vector in

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

Given that there exists a vector

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Let ∧u=u1∧i+u2∧j+u3∧k be a unit vector in R3 and ∧w=1√6(∧i+∧j+2∧k). Given that there exists a vector →v in R3 such that ∣∣∧u×→v∣∣=1 and ∧w.(∧u×→v)=1. Which of the following statements is(are) correct?