Let a unit vector u = x i + y j + z k make angles 2, 3, 2 3 with the vectors p1 = 1 2 i + 1 2 k, p2 = 1 2 j + 1 2 k, and p3 = 1 2 i + 1 2 j, respectively. If v = 1 2( i + j + k), then |u - v|^2 is…

Given: u = x i + y j + z k is a unit vector, and v = 1 2( i + j + k). Find: |u-v|^2. Use the dot product relation for angle between vectors: = a b| a| | b| Since the given vectors p1, p2, p3 are unit vectors, the dot products directly give the corresponding cosines. From the angle 2 with p1 = 1 2 i + 1 2 k, u p1 = 0 so x 2 + z 2 = 0 Hence, x = -z From the angle 3 with p2 = 1 2 j + 1 2 k, u p2 = 12 so y 2 + z 2 = 12 Hence, y+z = 22 From the angle 2 3 with p3 = 1 2 i + 1 2 j, u p3 = - 12 so x 2 + y 2 = - 12 Hence, x+y = - 22 Now use x=-z in the other two equations: y+z = 22 -z+y = - 22 Adding, 2y = 0 y = 0 Then, z = 22, x = - 22 Now compute |u-v|^2 = (x- 1 2)^2 + (y- 1 2)^2 + (z- 1 2)^2 Substituting x=- 1 2, y=0, z= 1 2, |u-v|^2 = (- 1 2- 1 2)^2 + (0- 1 2)^2 + ( 1 2- 1 2)^2 = (- 2)^2 + 12 + 0 = 2 + 12 = 52 Therefore, |u-v|^2 = 52, so the correct option is B. Given: u and v are vectors with |u|=1 and v = 1 2( i+ j+ k). Find: |u-v|^2. After obtaining x = - 1 2, y = 0, z = 1 2 we can use the identity |u-v|^2 = |u|^2 + |v|^2 - 2u v Now, |u|^2 = 1 and |v|^2 = ( 1 2)^2 + ( 1 2)^2 + ( 1 2)^2 = 32 Also, u v = (- 1 2) ( 1 2) + 0 1 2 + ( 1 2) ( 1 2) = - 12+0+ 12=0 Hence, |u-v|^2 = 1 + 32 - 2(0) = 52 Therefore, the correct option is B.

Let a unit vector u = x i + y j + z k make angles 2, 3, 2 3 with… - Solution | JEE Mathematics

MCQMediumJEE 2024Dot Product

Let a unit vector $u = x\hat{i} + y\hat{j} + z\hat{k}$ make angles $\frac{\pi}{2}$ , $\frac{\pi}{3}$ , $\frac{2\pi}{3}$ with the vectors $p_1 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$ , $p_2 = \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$ , and $p_3 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$ , respectively. If $v = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j} + \hat{k})$ , then $|u - v|^2$ is equal to:

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

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $u = x\hat{i} + y\hat{j} + z\hat{k}$ is a unit vector, and $v = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j} + \hat{k})$ .

Find: $|u-v|^2$ .

Use the dot product relation for angle between vectors:

\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}

Since the given vectors $p_1, p_2, p_3$ are unit vectors, the dot products directly give the corresponding cosines.

From the angle $\frac{\pi}{2}$ with $p_1 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$ ,

u \cdot p_1 = 0

\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0

Hence,

x = -z

From the angle $\frac{\pi}{3}$ with $p_2 = \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$ ,

u \cdot p_2 = \frac{1}{2}

\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2}

Hence,

y+z = \frac{\sqrt{2}}{2}

From the angle $\frac{2\pi}{3}$ with $p_3 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$ ,

u \cdot p_3 = -\frac{1}{2}

\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2}

Hence,

x+y = -\frac{\sqrt{2}}{2}

Now use $x=-z$ in the other two equations:

y+z = \frac{\sqrt{2}}{2}

-z+y = -\frac{\sqrt{2}}{2}

Adding,

2y = 0 \Rightarrow y = 0

Then,

z = \frac{\sqrt{2}}{2}, \qquad x = -\frac{\sqrt{2}}{2}

Now compute

|u-v|^2 = \left(x-\frac{1}{\sqrt{2}}\right)^2 + \left(y-\frac{1}{\sqrt{2}}\right)^2 + \left(z-\frac{1}{\sqrt{2}}\right)^2

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

|u-v|^2 = \left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)^2 + \left(0-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)^2

= \left(-\sqrt{2}\right)^2 + \frac{1}{2} + 0 = 2 + \frac{1}{2} = \frac{5}{2}

Therefore, $|u-v|^2 = \frac{5}{2}$ , so the correct option is B.

Using distance identity

Given: $u$ and $v$ are vectors with $|u|=1$ and $v = \frac{1}{\sqrt{2}}(\hat{i}+\hat{j}+\hat{k})$ .

Find: $|u-v|^2$ .

After obtaining

x = -\frac{1}{\sqrt{2}}, \qquad y = 0, \qquad z = \frac{1}{\sqrt{2}}

we can use the identity

|u-v|^2 = |u|^2 + |v|^2 - 2u\cdot v

Now,

|u|^2 = 1

and

|v|^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{3}{2}

Also,

u\cdot v = \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) + 0\cdot \frac{1}{\sqrt{2}} + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2}+0+\frac{1}{2}=0

Hence,

|u-v|^2 = 1 + \frac{3}{2} - 2(0) = \frac{5}{2}

Therefore, the correct option is B.

Common mistakes

Using the wrong vector with the wrong angle is a common mistake. Each angle must be matched with its corresponding vector $p_1, p_2, p_3$ exactly as given. Reassigning them changes the system of equations completely.
Forgetting that the given vectors $p_1, p_2, p_3$ are unit vectors leads to incorrect dot-product equations. Since their magnitudes are $1$ , the dot product equals the cosine directly.
Sign errors in cosine values are frequent. In particular, $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$ , not $\frac{1}{2}$ . This negative sign is essential for obtaining the correct values of $x, y, z$ .

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