Question:
For \(p \gt 0\), a vector
\(\vec{v}_2 = 2\hat{i} + (p+1)\hat{j}\)
is obtained by rotating the vector
\(\vec{v}_1 = \sqrt{3}\,p\hat{i} + \hat{j}\)
by an angle \(\theta\) about origin in counter clockwise direction. If
\(\tan\theta = \frac{(\alpha\sqrt{3} - 2)}{(4\sqrt{3} + 3)}\)
, then the value of \(\alpha\) is equal to ______________.
For \(p \gt 0\), a vector
\(\vec{v}_2 = 2\hat{i} + (p+1)\hat{j}\)
is obtained by rotating the vector
\(\vec{v}_1 = \sqrt{3}\,p\hat{i} + \hat{j}\)
by an angle \(\theta\) about origin in counter clockwise direction. If
\(\tan\theta = \frac{(\alpha\sqrt{3} - 2)}{(4\sqrt{3} + 3)}\)
, then the value of \(\alpha\) is equal to ______________.
\(\vec{v}_2 = 2\hat{i} + (p+1)\hat{j}\)
is obtained by rotating the vector
\(\vec{v}_1 = \sqrt{3}\,p\hat{i} + \hat{j}\)
by an angle \(\theta\) about origin in counter clockwise direction. If
\(\tan\theta = \frac{(\alpha\sqrt{3} - 2)}{(4\sqrt{3} + 3)}\)
, then the value of \(\alpha\) is equal to ______________.
Show Hint
Rotation preserves length, so \(|\vec v_1| = |\vec v_2|\) fixes p; then \(\tan\theta\) follows from the cross and dot products of the two vectors.
Updated On: Nov 4, 2026
Show Solution
Verified By Collegedunia
Correct Answer: 6
Solution and Explanation
Rotation keeps the magnitude: \(3p^2 + 1 = 4 + (p+1)^2 \Rightarrow 2p^2 - 2p - 4 = 0 \Rightarrow p = 2\) (p > 0).
So \(\vec v_1 = (2\sqrt3, 1)\), \(\vec v_2 = (2, 3)\).
Slopes: \(m_1 = \frac{1}{2\sqrt3}\), \(m_2 = \frac{3}{2}\).
\(\tan\theta = \frac{m_2 - m_1}{1 + m_1 m_2} = \frac{\frac{3}{2} - \frac{1}{2\sqrt3}}{1 + \frac{3}{4\sqrt3}} = \frac{(3\sqrt3 - 1)/(2\sqrt3)}{(4\sqrt3 + 3)/(4\sqrt3)} = \frac{6\sqrt3 - 2}{4\sqrt3 + 3}\).
Comparing with \(\frac{\alpha\sqrt3 - 2}{4\sqrt3 + 3}\) gives \(\alpha = 6\).
Answer: 6
So \(\vec v_1 = (2\sqrt3, 1)\), \(\vec v_2 = (2, 3)\).
Slopes: \(m_1 = \frac{1}{2\sqrt3}\), \(m_2 = \frac{3}{2}\).
\(\tan\theta = \frac{m_2 - m_1}{1 + m_1 m_2} = \frac{\frac{3}{2} - \frac{1}{2\sqrt3}}{1 + \frac{3}{4\sqrt3}} = \frac{(3\sqrt3 - 1)/(2\sqrt3)}{(4\sqrt3 + 3)/(4\sqrt3)} = \frac{6\sqrt3 - 2}{4\sqrt3 + 3}\).
Comparing with \(\frac{\alpha\sqrt3 - 2}{4\sqrt3 + 3}\) gives \(\alpha = 6\).
Answer: 6
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Vector Algebra Questions
- Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \( |\vec{b}| = 1 \) and \( |\vec{b} \times \vec{a}| = 2 \). Then \( \left| (\vec{b} \times \vec{a}) - \vec{b} \right|^2 \) is equal to
- The position vectors of the vertices \( A, B \) and \( C \) of a triangle are \[ 2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}, \quad 2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \quad \text{and} \quad -\mathbf{i} + \mathbf{j} + 3\mathbf{k} \] respectively. Let \( l \) denote the length of the angle bisector \( AD \) of \( \angle BAC \) where \( D \) is on the line segment \( BC \). Then \( 2l^2 \) equals:
- Let the position vectors of the vertices \( A, B \) and \( C \) of a triangle be \[ 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}, \quad \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \quad \text{and} \quad 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \] respectively. Let \( l_1, l_2 \) and \( l_3 \) be the lengths of the perpendiculars drawn from the ortho center of the triangle on the sides \( AB, BC \) and \( CA \) respectively. Then \( l_1^2 + l_2^2 + l_3^2 \) equals:
- Let \(\overrightarrow{a},\overrightarrow{b}\) and \(\overrightarrow{c}\) be three non-zero vectors such that \(\overrightarrow{b}\) and \(\overrightarrow{c}\) are non-collinear. If \(\overrightarrow{a}+5\overrightarrow{b}\) is collinear with \(\overrightarrow{c},\overrightarrow{b}+6\overrightarrow{c}\) is collinear with \(\overrightarrow{a}\) and \(\overrightarrow{a}+ α\overrightarrow{b} + β\overrightarrow{c} = 0\), then α + β is equal to
- Let O be the origin and the position vector of A and B be \(2\hat{i}+2\hat{j}+\hat{k}\) and \(2\hat{i}+4\hat{j}+4\hat{k}\) respectively. If the internal bisector of\(\angle AOB\) meets the line AB at C, then the length of OC is
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions