JEE 2023 — Mathematics PYQ

Humein $\vec{a}$ aur $\vec{c}$ diye gaye hain:

• $\vec{a}=3\hat{i}+\hat{j}-\hat{k}⟹|\vec{a}{|}^2=3^2+1^2+(-1{)}^2=9+1+1=11$

• $\vec{c}=2\hat{i}-3\hat{j}+3\hat{k}⟹|\vec{c}{|}^2=2^2+(-3{)}^2+3^2=4+9+9=22$

Step 2: $\vec{b}\cdot \vec{c}$ ki value nikalna

Humein diya gaya hai ki $\vec{a}=\vec{b}\times \vec{c}$.

Kyuki cross product hamesha dono vectors ke perpendicular hota hai, iska matlab $\vec{a}\perp \vec{b}$ aur $\vec{a}\perp \vec{c}$.

(Verify: $\vec{a}\cdot \vec{c}=(3)(2)+(1)(-3)+(-1)(3)=6-3-3=0$. Ye sahi hai.)

Ab cross product ka magnitude formula use karte hain:

Jahan $\theta$, $\vec{b}$ aur $\vec{c}$ ke beech ka angle hai.

Humein pata hai ki ${\sin }^2\theta =1-{\cos }^2\theta$:

Values rakhte hain:

Step 3: $|\vec{b}+\vec{c}{|}^2$ nikalna

Step 4: Final Answer

Humein $|72-|\vec{b}+\vec{c}{|}^2|$ ki value nikalni hai:

Case 1: Agar $|\vec{b}+\vec{c}{|}^2=72+66$

Case 2: Agar $|\vec{b}+\vec{c}{|}^2=72-66$

Dono cases mein answer same hai.

Final Answer:

66