Let , and . If is a vector such that , and the angle between and is , then is equal to __________. [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{d} = \vec{a} \times \vec{b}$ . If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} =$ $| \vec{c} |$ , $| \vec{c} - 2 \vec{a} |^{2} = 8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{π}{4}$ , then $| 10 - 3 \vec{b} \cdot \vec{c} |$ $+$ $| \vec{d} \times \vec{c} |^{2}$ is equal to __________. [2025]

Ans.
(6)

Given, $\vec{d} = \vec{a} \times \vec{b}$ , where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}$

$∴ \vec{d} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{matrix} |$ $= - \hat{i} + \hat{j}$ ... (i)

Also, $| \vec{c} - 2 \vec{a} |^{2} = 8 \Rightarrow | \vec{c} |^{2} + 4 | \vec{a} |^{2} - 4 \vec{a} \cdot \vec{c} = 8$

$\Rightarrow | \vec{c} |^{2} + 4 (3) - 4 | \vec{c} | = 8$ [Given, $\vec{a} \cdot \vec{c} =$ $| \vec{c} |$ ]

$\Rightarrow | \vec{c} |^{2} - 4 | \vec{c} | + 4 = 0 \Rightarrow {(| \vec{c} | - 2)}^{2} = 0 \Rightarrow | \vec{c} | = 2$

Now, $\vec{d} = \vec{a} \times \vec{b} \Rightarrow$ $| \vec{d} \times \vec{c} |$ $=$ $| (\vec{a} \times \vec{b}) \times \vec{c} |$

Using triple vector product properties, we have

$| \vec{d} |$ $| \vec{c} |$ $\sin \frac{π}{4} =$ $| ((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}) |$

$\sqrt{2} \times 2 \times \frac{1}{\sqrt{2}}$ $| 2 \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} |$

Squaring both sides, we get

$4 = 4 | \vec{b} |^{2} + {(\vec{b} \cdot \vec{c})}^{2} | \vec{a} |^{2} - 4 (\vec{b} \cdot \vec{c}) (\vec{a} \cdot \vec{b})$

$\Rightarrow 4 = 4 (9) + 3 {(\vec{b} \cdot \vec{c})}^{2} - 4 \times 5 (\vec{b} \cdot \vec{c})$

$\Rightarrow 3 {(\vec{b} \cdot \vec{c})}^{2} - 20 {(\vec{b} \cdot \vec{c})}^{2} + 32 = 0$

Let $\vec{b} \cdot \vec{c} = t$ , we have

$3 t^{2} - 20 t + 32 = 0 \Rightarrow t = \frac{8}{3}, 4 \Rightarrow \vec{b} \cdot \vec{c} = \frac{8}{3} or \vec{b} \cdot \vec{c} = 4$

$∴$ $| 10 - 3 \vec{b} \cdot \vec{c} |$ $+ | \vec{d} \times \vec{c} |^{2} =$ $| 10 - 3 (\frac{8}{3}) |$ $+ {(2)}^{2} = 6$

or $| 10 - 3 (4) |$ $+ {(2)}^{2} = 6$ .

Want Us to Email you About Special Offers & Updates?