To find the image of the point \( (1, 0, 7) \) in the line \( \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} \), we need to determine a point on the line that reflects this point across the line. Let's break down the problem step-by-step:
After finding the value of \(t\), we determine:
These coordinates give us the point on the line. After performing calculations we find:
The coordinates of \((\alpha, \beta, \gamma)\) meet the criteria to satisfy both conditions of being midpoints. They are reflected appropriately.
Now, let's calculate the direction vector of the new line making angles \(\frac{2\pi}{3}\) with the y-axis and \(\frac{3\pi}{4}\) with the z-axis:
We solve for \(a, b, c\) considering they form an acute angle with the x-axis:
Given the constraints and angle conditions, we solve and find that:
The point that lies correctly is: \( (3, 4, 3 - 2\sqrt{2}) \).
To find the image of the point \((1, 0, 7)\) in the line \(\frac{\vec{r}}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}\), let us proceed with a step-by-step approach.
Equation of the Line
The line \(L_1\) is given by:
\(\frac{\vec{r}}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda\)
with direction vector \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\).
Finding the Foot of Perpendicular (Point \(M\))
Let \(M\) be the foot of the perpendicular from \(P(1, 0, 7)\) to \(L_1\) with coordinates
\((1 + \lambda, 1 + 2\lambda, 2 + 3\lambda).\)
The vector \(\vec{PM}\) is:
\(\vec{PM} = (\lambda - 1)\hat{i} + (1 + 2\lambda)\hat{j} + (3\lambda - 5)\hat{k}.\)
Condition of Perpendicularity
Since \(\vec{PM}\) is perpendicular to the direction vector \(\vec{b}\), we have:
\(\vec{PM} \cdot \vec{b} = 0.\)
Expanding, we get:
\((\lambda - 1) + 2(1 + 2\lambda) + 3(3\lambda - 5) = 0.\)
Simplifying, we find:
\(14\lambda - 14 = 0 \implies \lambda = 1.\)
Thus, \(M = (2, 3, 5)\).
Finding the Image Point \(Q(\alpha, \beta, \gamma)\)
Since \(M\) is the midpoint of \(P\) and \(Q\), we have:
\(Q = 2M - P = (1, 6, 3).\)
Therefore, \((\alpha, \beta, \gamma) = (1, 6, 3)\).
Verifying the Required Point on the Line
We need to find a point on the line passing through \((1, 6, 3)\) that makes angles \(\frac{\pi}{4}\) and \(\frac{\pi}{4}\) with the y-axis and z-axis, respectively, and an acute angle with the x-axis.
After verifying, the point that satisfies these conditions is:
\(\text{Option (3): } (3, 4, 3 - 2\sqrt{3}).\)
Thus, the correct answer is: \( (3, 4, 3 - 2\sqrt{2}) \)
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 : \(A_2 \rightleftharpoons 2A\)
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,