A vector has magnitude same as that of A → = 3 i ^ + 4 j ^ and is parallel to B → = 4 i ^ + 3 j ^ . The x and y components of this vector in first quadrant are x and 3 respectively where x = ____.

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Aicharya · Accepted Answer

Magnitude of vector A → = 3 2 + 4 2 = 5 Now, unit vector along B → = ( 4 i ^ + 3 j ^ ) 4 2 + 3 2 = 4 5 i ^ + 3 5 j ^ Now, N → = A → B ^ = 5 × 4 5 i ^ + 3 5 j ^ = 4 i ^ + 3 j ^ ∴ x = 4 Alternative Solution: Arey, dekho! Vectors ke sawal JEE mein bahut common hote hain aur yeh wala toh bohot hi simple concept par based hai: agar tumhe kisi vector ka **magnitude** aur **direction** pata hai, toh tum us vector ko easily likh sakte ho. Is sawal ko hum step-by-step tod kar solve karenge. ### Step 1: Required Magnitude Calculate Karna Humein jo final vector chahiye, chalo usko $\vec{C}$ maan lete hain. Condition diya hai ki $|\vec{C}|$ ka magnitude utna hi hai jitna $\vec{A}$ ka hai. Yahaan, ek chhota sa typo lag raha hai, $\vec{A}=3\hat{j}+4\hat{j}$ toh $\vec{A}=7\hat{j}$ ho jayega. Hum assume karte hain ki original intent $\vec{A}=3\hat{i}+4\hat{j}$ tha, jo standard JEE questions mein hota hai. Let's take $\vec{A} = 3\hat{i} + 4\hat{j}$. Magnitude of $\vec{A}$ kya hota hai? Pythagoras theorem use karo: $$|\vec{A}| = \sqrt{(A_x)^2 + (A_y)^2}$$ $$|\vec{A}| = \sqrt{3^2 + 4^2}$$ $$|\vec{A}| = \sqrt{9 + 16} = \sqrt{25}$$ $$|\vec{A}| = 5$$ Toh, hamare required vector $\vec{C}$ ka magnitude kitna ho gaya? $$|\vec{C}| = 5$$ Bas, itna toh easy tha! *** ### Step 2: Required Direction Determine Karna Ab humein $\vec{C}$ ka direction chahiye. Direction kis se pata chalti hai? Unit Vector se. Condition kya hai? $\vec{C}$ is parallel to $\vec{B} = 4\hat{i} + 3\hat{j}$. Agar do vectors parallel hote hain, toh unka unit vector same hota hai. Pehle $\vec{B}$ ka magnitude nikalte hain: $$|\vec{B}| = \sqrt{4^2 + 3^2}$$ $$|\vec{B}| = \sqrt{16 + 9} = \sqrt{25}$$ $$|\vec{B}| = 5$$ Ab $\vec{B}$ ka unit vector $\hat{B}$ nikalte hain. Yeh hamare $\vec{C}$ ki direction hogi: $$\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{4\hat{i} + 3\hat{j}}{5}$$ *** ### Step 3: Final Vector $\vec{C}$ Construct Karna Humein magnitude ($|\vec{C}| = 5$) aur direction ($\hat{B}$) dono mil gaye. Koi bhi vector kya hota hai? **Magnitude $	imes$ Direction**. $$\vec{C} = |\vec{C}| \cdot \hat{B}$$ $$\vec{C} = 5 \cdot \left(\frac{4\hat{i} + 3\hat{j}}{5}ight)$$ Dekho, 5 aur 5 cancel ho jayenge! $$\vec{C} = 4\hat{i} + 3\hat{j}$$ *** ### Step 4: Components Compare Karke $x$ Find Karna Sawal mein diya gaya hai ki $\vec{C}$ ke $x$ aur $y$ components $x$ aur $3$ hain, respectively. Matlab, $$\vec{C} = x\hat{i} + 3\hat{j}$$ Ab hum Step 3 aur Step 4 ke results ko compare karte hain: $$4\hat{i} + 3\hat{j} = x\hat{i} + 3\hat{j}$$ Components compare karo: 1. $y$-component: $3 = 3$ (Correct) 2. $x$-component: $x = 4$ Toh, final answer $x=4$ hai. Samjha? Vector ka sawal jab bhi aaye, bas yahi funda yaad rakhna: **Vector = Scalar (Magnitude) times Unit Vector (Direction)**.

A vector has magnitude same as that of A → = 3 i ^ + 4 j ^ and is parallel to B → = 4 i ^ + 3 j ^ . The x…

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