Let v- be a vector in the plane such that -v- - i- - -v- - 2-i- - -v- - j- Then -v- lies in the interval-

The correct option is A (1, 2) Given #160; | v⃗ i⃗ |= | v⃗ 2i⃗ |= | v⃗ j⃗ | taking | v⃗ i⃗ |= | v⃗ 2i⃗ | on squaring both sides #160; | v⃗ ...

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Byju's Answer

Standard XII

Mathematics

Combination with Restrictions

Let v⃗ be a...

Question

Let $\to v$ be a vector in the plane such that $| \to v - \to i | = | \to v - \to 2 i | = | \to v - \to j |$ . Then $| \to v |$ lies in the interval.

(0, 1)

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(1, 2)

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(2, 3)

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(3, 4)

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Solution

The correct option is A $(1, 2)$
Given
$∣ ∣ \to v - \to i ∣ ∣ = ∣ ∣ \to v - 2 \to i ∣ ∣ = ∣ ∣ \to v - \to j ∣ ∣$
taking
$∣ ∣ \to v - \to i ∣ ∣ = ∣ ∣ \to v - 2 \to i ∣ ∣$
on squaring both sides
${| \to v |}^{2} + {∣ ∣ \to i ∣ ∣}^{2} - 2 \to v \cdot \to i = {| \to v |}^{2} + {∣ ∣ 2 \to i ∣ ∣}^{2} - 4 \to v \cdot \to i$
${| \to v |}^{2} - {| \to v |}^{2} - 4 {∣ ∣ \to i ∣ ∣}^{2} + {∣ ∣ \to i ∣ ∣}^{2} - 2 \to v \cdot \to i = - 4 \to v \cdot \to i$
$- 3 {∣ ∣ \to i ∣ ∣}^{2} = - 4 \to v \cdot \to i + 2 \to v \cdot \to i$
$- 3 {∣ ∣ \to i ∣ ∣}^{2} = - 2 \to v \cdot \to i$
$\to v$ is in th plane so it is parallel to i
$- 3 = - 2 | \to v | ∣ ∣ \to i ∣ ∣$
$| \to v | = \frac{3}{2}$
So the value is betweem 2 and 3

Suggest Corrections

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Let →v be a vector in the plane such that |→v−→i|=|→v−→2i|=|→v−→j|. Then |→v| lies in the interval.

Let $\to v$ be a vector in the plane such that $| \to v - \to i | = | \to v - \to 2 i | = | \to v - \to j |$ . Then $| \to v |$ lies in the interval.