Each of the angles \( \beta \) and \( \gamma \) that a given line makes with the positive y- and z-axes, respectively, is half the angle that this line makes with the positive x-axis. Then the sum of all possible values of the angle \( \beta \) is
Show Hint
- \( \frac{3\pi}{4} \)
- \( \pi \)
- \( \frac{\pi}{2} \)
- \( \frac{3\pi}{2} \)
The Correct Option is A
Approach Solution - 1
To solve the problem, let's consider the given conditions regarding the angles a line makes with the axes. Denoting large capital angles: \( \alpha \) with the x-axis, \( \beta \) with the y-axis, and \( \gamma \) with the z-axis.
According to the problem statement:
- \( \beta = \frac{\alpha}{2} \)
- \( \gamma = \frac{\alpha}{2} \)
- We need to find the sum of all possible values of the angle \( \beta \).
The direction cosines of the line with respect to the x-, y-, and z-axes are given by:
- \(\cos \alpha\), \(\cos \beta\), \(\cos \gamma\)
According to the identity involving direction cosines, we have:
\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)Substitute the known relations into the equation:
- \(\cos^2 \alpha + \left(\cos \frac{\alpha}{2}\right)^2 + \left(\cos \frac{\alpha}{2}\right)^2 = 1\)
This simplifies to:
\(\cos^2 \alpha + 2 \cos^2 \frac{\alpha}{2} = 1\)Using the double angle identity: \( \cos^2 \frac{\alpha}{2} = \frac{1 + \cos \alpha}{2} \), we substitute and get:
\(\cos^2 \alpha + 2 \times \frac{1 + \cos \alpha}{2} = 1\)This further simplifies to:
\(\cos^2 \alpha + 1 + \cos \alpha = 1\)Or, reorganizing terms:
\(\cos^2 \alpha + \cos \alpha = 0\)Factor the quadratic equation:
\(\cos \alpha (\cos \alpha + 1) = 0\)This gives us two solutions:
- \( \cos \alpha = 0 \Rightarrow \alpha = \frac{\pi}{2} \)
- \( \cos \alpha = -1 \Rightarrow \alpha = \pi \)
Substituting back, we find the possible values of \( \beta \):
- For \(\alpha = \frac{\pi}{2}\):
- \(\beta = \frac{\pi}{4}\)
- For \(\alpha = \pi\):
- \(\beta = \frac{\pi}{2}\)
The sum of all possible values of \( \beta \) is then:
\(\frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}\)Therefore, the sum of all possible values of the angle \( \beta \) is \(\frac{3\pi}{4}\).
Approach Solution -2
Let the angle with the positive x-axis be \( \alpha \).
\(\text{Given, } \beta = \frac{\alpha}{2} \text{ and } \gamma = \frac{\alpha}{2}. \)
\(\text{We know that } \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. \)
\(\text{Substituting the values of } \beta \text{ and } \gamma: \)
\(\cos^2 \alpha + \cos^2 \left( \frac{\alpha}{2} \right) + \cos^2 \left( \frac{\alpha}{2} \right) = 1 \)
\(\Rightarrow \cos^2 \alpha + 2 \cos^2 \left( \frac{\alpha}{2} \right) = 1 \)
\(\text{Using the identity } \cos \alpha = 2 \cos^2 \left( \frac{\alpha}{2} \right) - 1, \text{ we get } \)
\(2 \cos^2 \left( \frac{\alpha}{2} \right) = \cos \alpha + 1 \)
\(\text{So,} \)
\(\cos^2 \alpha + \cos \alpha + 1 = 1 \)
\(\Rightarrow \cos^2 \alpha + \cos \alpha = 0 \)
\(\Rightarrow \cos \alpha (\cos \alpha + 1) = 0 \)
\(\text{This gives } \cos \alpha = 0 \text{ or } \cos \alpha = -1. \)
Case 1:
\(\cos \alpha = 0 \)
\(\Rightarrow \alpha = \frac{\pi}{2} \text{ or } \alpha = \frac{3\pi}{2} \)
\(\text{Since the angles are with the positive axes, } 0 \le \alpha, \beta, \gamma \le \pi. \)
\(\text{If } \alpha = \frac{\pi}{2}, \text{ then } \beta = \frac{\pi}{4} \)
\(\text{If } \alpha = \frac{3\pi}{2}, \text{ this is not possible as } \beta = \frac{3\pi}{4} \text{ and } \gamma = \frac{3\pi}{4}, \)
\(\text{leading to } \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 0 + \frac{1}{2} + \frac{1}{2} = 1. \)
Case 2:
\(\cos \alpha = -1 \)
\(\Rightarrow \alpha = \pi \)
\(\Rightarrow \beta = \frac{\pi}{2}, \quad \gamma = \frac{\pi}{2} \)
\(\Rightarrow \cos^2 \pi + \cos^2 \frac{\pi}{2} + \cos^2 \frac{\pi}{2} = 1 + 0 + 0 = 1 \)
\(\text{Possible values of } \beta \text{ are } \frac{\pi}{4} \text{ and } \frac{\pi}{2}. \)
\(\text{Sum of possible values of } \beta = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}. \)
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_____
Top JEE Main 3D Geometry Questions
- Let line L pass through the point \( (0, 1, 2) \), intersect the line \( {2 = {3 = {4 \), and be parallel to the plane \( 2x + y - 3z = 4 \). Find the distance of the point \( P(1, -9, 2) \) from line L. \\
- Let the vectors \( , , \) represent three coterminous edges of a parallelepiped of volume \( V \). Then the volume of the parallelepiped, whose coterminous edges are represented by \( , + \) and \( + 2 + 3 \) is equal to: \\ \\
- If \( d_1 \) is the shortest distance between the lines \(\frac{x+1}{2} = \frac{y-1}{-12} = \frac{z-1}{-7} \quad \text{and} \quad \frac{x-1}{7} = \frac{y+8}{2} = \frac{z-4}{5},\) and \( d_2 \) is the shortest distance between the lines \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3} \quad \text{and} \quad \frac{x+2}{1} = \frac{y+2}{2} = \frac{z-1}{6},\) then the value of \( \frac{32\sqrt{3}d_1}{d_2} \) is:
- The lines \[ \frac{x - 2}{2} = \frac{y + 2}{-2} = \frac{z - 7}{16} \] and \[ \frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1} \] intersect at the point \( P \). If the distance of \( P \) from the line \[ \frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1} \] is \( l \), then \( 14l^2 \) is equal to \ldots
- Let PQR be a triangle with R (-1,4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of \(\triangle PQR\) from the point of intersection of the line\(\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}\) and \(\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}\) is
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,