For p > 0, the Vector v₂ = 2i + (p+1)j Is Obtained by Rotating v₁ = √3 p i + j by an Angle θ About the Origin in a Counterclockwise Direction. If tan θ = (α√3 − 2)/(4√3 + 3), Then the Value of α Is Equal To

JEE Main 2021. Previous Year Question.

Problem

For $p>0$, the vector $\vec{v_2}=2\hat i+(p+1)\hat j$ is obtained by rotating the vector $\vec{v_1}=\sqrt3\,p\,\hat i+\hat j$ by an angle $\theta$ about the origin in a counterclockwise direction. If $\tan\theta=\dfrac{\alpha\sqrt3-2}{4\sqrt3+3}$, then the value of $\alpha$ is equal to:

Key insight. A rotation never changes a vector’s length — so $|\vec{v_1}|=|\vec{v_2}|$ is a genuine equation that pins down $p$ before anything else is computed. Only after $p$ is known does it make sense to compute $\tan\theta$ from the standard “cross over dot” formula for the angle between two vectors.

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Approach

Use the fact that rotation preserves length to set $|\vec{v_1}|=|\vec{v_2}|$ and solve for $p$ (taking the positive root, since $p>0$). With $p$ known, compute $\tan\theta$ between $\vec{v_1}$ and $\vec{v_2}$ using $\tan\theta=\dfrac{\vec{v_1}\times\vec{v_2}}{\vec{v_1}\cdot\vec{v_2}}$ (the $z$-component of the cross product over the dot product), then match it against the given form to read off $\alpha$.

Solution

Step 1 — Set |v1| = |v2| since rotation preserves length

$$3p^2+1 = 4+(p+1)^2 = p^2+2p+5$$

Step 2 — Solve for p

$$2p^2-2p-4=0 \implies p^2-p-2=0 \implies (p-2)(p+1)=0$$

Since $p>0$: $p=2$.

Step 3 — Substitute p = 2 into both vectors

$$\vec{v_1} = 2\sqrt3\,\hat i+\hat j, \qquad \vec{v_2} = 2\hat i+3\hat j$$

Step 4 — Compute the cross product’s z-component

$$\vec{v_1}\times\vec{v_2}\big|_z = (2\sqrt3)(3)-(1)(2) = 6\sqrt3-2$$

Step 5 — Compute the dot product

$$\vec{v_1}\cdot\vec{v_2} = (2\sqrt3)(2)+(1)(3) = 4\sqrt3+3$$

Step 6 — Form tan θ

$$\tan\theta = \frac{6\sqrt3-2}{4\sqrt3+3}$$

Step 7 — Match against the given form to find α

Comparing $\dfrac{\alpha\sqrt3-2}{4\sqrt3+3}$ with $\dfrac{6\sqrt3-2}{4\sqrt3+3}$: the denominators already match exactly, and equating numerators gives $\alpha\sqrt3=6\sqrt3$, so:

$$\alpha=6$$

Answer

Common mistakes

• Computing tan θ before finding p. Since $\vec{v_1}$ and $\vec{v_2}$ both contain $p$, the cross and dot products only simplify to match the given form of $\tan\theta$ once the specific value of $p$ is substituted in — attempting to match symbolically in $p$ first leads nowhere.
• Taking the negative root $p=-1$. The quadratic $p^2-p-2=0$ has two roots, but the problem explicitly restricts to $p>0$, so only $p=2$ is valid.

Practise next

• For $p>0$, a vector $\vec{v_2}=3\hat i+(p+2)\hat j$ is obtained by rotating $\vec{v_1}=2p\hat i+\hat j$ counterclockwise by $\theta$. If $\tan\theta=\dfrac{\alpha-3}{2p+3}$ for the resulting $p$, find $\alpha$, using the same length-preservation approach.