Let a unit vector ^ u = x ^ i + y ^ j + z ^ k make angles (π)/(2), (π)/(3) and (2π)/(3) with the vectors 1 2 ^ i + 1 2 ^ k , 1 2 ^ j + 1 2 ^ k and 1 2 ^ i + 1 2 ^ j respectively. If → v = 1 2 ^ i + 1 2 ^ j + 1 2 ^ k

Correct answer: B. (5)/(2) Unit vector ^ u = x ^ i + y ^ j + z ^ k → p 1 = 1 2 ^ i + 1 2 ^ k , → p 2 = 1 2 ^ j + 1 2 ^ k → p 3 = 1 2 ^ i + 1 2 ^ j Now angle between ^ u and → p 1 = (π)/(2) ^ u · → p 1 = 0 ⇒ x 2 + z 2 = 0 ⇒ x + z = 0 (i) Angle between ^ u and → p 2 = (π)/(3) ^ u · → p 2 = | ^ u | · | → p 2|cos(π)/(3) ^ u · → p 2 = y 2 + z 2 = (1)/(2) (ii) Angle between ^ u and → p 3 = (2π)/(3) ^ u · → p 3 = | ^ u | · | → p 3|cos(2π)/(3) ⇒ x 2 + y 2 = (- 1)/(2) ⇒ x + y = - 1 2 (iii) from equation (i), (ii) and (iii) we get x = - 1 2 , y = 0, z = 1 2 Thus ^ u - → v = - 1 2 ^ i + 1 2 ^ k - 1 2 ^ i - 1 2 ^ j - 1 2 ^ k ⇒ ^ u - → v = - 2 2 ^ i - 1 2 ^ j ∴ | ^ u - → v | ^ 2 = ( (4)/(2) + (1)/(2) ) ^ 2 = (5)/(2)

JEE Main 202429 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2024 · 29 January · Evening Shift · Mathematics · Vector Algebra

Let a unit vector ^ u = x ^ i + y ^ j + z ^ k make angles (π)/(2),...

Let a unit vector

\overset{^}{u} = x\overset{^}{i} + y\overset{^}{j} + z\overset{^}{k}

make angles

\dfrac{\pi}{2}, \dfrac{\pi}{3}

and

\dfrac{2\pi}{3}

with the vectors

\dfrac{1}{\sqrt{2}}\overset{^}{i} + \dfrac{1}{\sqrt{2}}\overset{^}{k}, \dfrac{1}{\sqrt{2}}\overset{^}{j} + \dfrac{1}{\sqrt{2}}\overset{^}{k}

and

\dfrac{1}{\sqrt{2}}\overset{^}{i} + \dfrac{1}{\sqrt{2}}\overset{^}{j}

respectively. If

\overset{\to}{v} = \dfrac{1}{\sqrt{2}}\overset{^}{i} + \dfrac{1}{\sqrt{2}}\overset{^}{j} + \dfrac{1}{\sqrt{2}}\overset{^}{k}

, then (|\hat{u}-\vec{v}|^2) is equal to

Options

A $\dfrac{11}{2}$
B $\dfrac{5}{2}$
C $9$
D $7$

Correct answer

B. $\dfrac{5}{2}$

Step-by-step solution

Unit vector $\overset{^}{u} = x\overset{^}{i} + y\overset{^}{j} + z\overset{^}{k}$

$\overset{\to}{p}1 = \dfrac{1}{\sqrt{2}}\overset{^}{i} + \dfrac{1}{\sqrt{2}}\overset{^}{k}, \overset{\to}{p}2 = \dfrac{1}{\sqrt{2}}\overset{^}{j} + \dfrac{1}{\sqrt{2}}\overset{^}{k}$

$\overset{\to}{p}3 = \dfrac{1}{\sqrt{2}}\overset{^}{i} + \dfrac{1}{\sqrt{2}}\overset{^}{j}$

Now angle between $\overset{^}{u}$ and $\overset{\to}{p}1 = \dfrac{\pi}{2}$

$\overset{^}{u} \cdot \overset{\to}{p}1 = 0 ⇒ \dfrac{x}{\sqrt{2}} + \dfrac{z}{\sqrt{2}} = 0$

$⇒ x + z = 0 \ldots \left(i\right)$

Angle between $\overset{^}{u}$ and $\overset{\to}{p}2 = \dfrac{\pi}{3}$

$\overset{^}{u} \cdot \overset{\to}{p}2 = | \overset{^}{u} | \cdot \left|\overset{\to}{p}2\right|cos\dfrac{\pi}{3}$

$\overset{^}{u} \cdot \overset{\to}{p}2 = \dfrac{y}{\sqrt{2}} + \dfrac{z}{\sqrt{2}} = \dfrac{1}{2} \ldots \left(ii\right)$

Angle between $\overset{^}{u}$ and $\overset{\to}{p}3 = \dfrac{2\pi}{3}$

$\overset{^}{u} \cdot \overset{\to}{p}3 = | \overset{^}{u} | \cdot \left|\overset{\to}{p}3\right|cos\dfrac{2\pi}{3}$

$⇒ \dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = \dfrac{- 1}{2} ⇒ x + y = \dfrac{- 1}{\sqrt{2}} \ldots \left(iii\right)$

from equation (i), (ii) and (iii) we get

$x = \dfrac{- 1}{\sqrt{2}}, y = 0, z = \dfrac{1}{\sqrt{2}}$

Thus $\overset{^}{u} - \overset{\to}{v} = \dfrac{- 1}{\sqrt{2}}\overset{^}{i} + \dfrac{1}{\sqrt{2}}\overset{^}{k} - \dfrac{1}{\sqrt{2}}\overset{^}{i} - \dfrac{1}{\sqrt{2}}\overset{^}{j} - \dfrac{1}{\sqrt{2}}\overset{^}{k}$

$⇒ \overset{^}{u} - \overset{\to}{v} = \dfrac{- 2}{\sqrt{2}}\overset{^}{i} - \dfrac{1}{\sqrt{2}}\overset{^}{j}$

$∴ | \overset{^}{u} - \overset{\to}{v}{|}^{2} = {\left(\sqrt{\dfrac{4}{2} + \dfrac{1}{2}}\right)}^{2} = \dfrac{5}{2}$

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