Question ID: #977
If the distance of the point $P(43, \alpha, \beta)$, $\beta<0$, from the line $\vec{r}=4\hat{i}-\hat{k}+\mu(2\hat{i}+3\hat{k})$, $\mu\in \mathbb{R}$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to ________.
Solution:
Equation of the given line $L_1$ in Cartesian form:
$$\frac{x – 4}{2} = \frac{y – 0}{0} = \frac{z – (-1)}{3} = \mu$$
Any point on this line $L_1$ is $P_1(2\mu + 4, 0, 3\mu – 1)$.
Equation of the line $L_2$ passing through $P(43, \alpha, \beta)$ with direction ratios $3, -1, 0$:
$$\frac{x – 43}{3} = \frac{y – \alpha}{-1} = \frac{z – \beta}{0} = \lambda$$
[span_2](start_span)Any point on this line $L_2$ is $P_1(3\lambda + 43, -\lambda + \alpha, \beta)$.[span_2](end_span)
Since the distance is measured along $L_2$ to the line $L_1$, the point $P_1$ must be the intersection of the two lines.
Equate the coordinates of $P_1$:
$$2\mu + 4 = 3\lambda + 43 \Rightarrow 2\mu – 3\lambda = 39 \quad \dots(1)$$
$$0 = -\lambda + \alpha \Rightarrow \lambda = \alpha \quad \dots(2)$$
$$3\mu – 1 = \beta \Rightarrow 3\mu = \beta + 1 \quad \dots(3)$$
Distance formula $d^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2$.
The distance between $P(43, \alpha, \beta)$ and $P_1(3\lambda + 43, -\lambda + \alpha, \beta)$ is $13\sqrt{10}$.
$$(3\lambda + 43 – 43)^2 + (-\lambda + \alpha – \alpha)^2 + (\beta – \beta)^2 = (13\sqrt{10})^2$$
$$(3\lambda)^2 + (-\lambda)^2 + 0 = 1690$$
$$9\lambda^2 + \lambda^2 = 1690$$
$$10\lambda^2 = 1690$$
$$\lambda^2 = 169 \Rightarrow \lambda = \pm 13$$
Since $\alpha = \lambda$, we have $\alpha = 13$ or $\alpha = -13$.
From equation (1), find $\mu$:
$$2\mu = 3\lambda + 39 \Rightarrow \mu = \frac{3\lambda + 39}{2}$$
From equation (3), substitute $\mu$ to find $\beta$:
$$\beta = 3\left(\frac{3\lambda + 39}{2}\right) – 1$$
$$\beta = \frac{9\lambda + 117 – 2}{2} = \frac{9\lambda + 115}{2}$$
Check the value of $\beta$ for both values of $\lambda$:
If $\lambda = 13$:
$$\beta = \frac{9(13) + 115}{2} = \frac{117 + 115}{2} = \frac{232}{2} = 116$$
This is rejected because given $\beta < 0$.
If $\lambda = -13$:
$$\beta = \frac{9(-13) + 115}{2} = \frac{-117 + 115}{2} = \frac{-2}{2} = -1$$
This is accepted since $\beta < 0$.
Therefore, $\alpha = -13$ and $\beta = -1$.
Calculate $\alpha^2 + \beta^2$:
$$\alpha^2 + \beta^2 = (-13)^2 + (-1)^2$$
$$\alpha^2 + \beta^2 = 169 + 1 = 170$$
Ans. 170
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