If the distance of the point P(43, alpha, bet | vedclass

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(A) The line is given by $\vec{r} = (4, 0, -1) + \mu(2, 0, 3)$. Let the point $P$ be $(43, \alpha, \beta)$. The line passing through $P$ with directio

Vedclass · Accepted Answer

$170$

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(A) The line is given by $\vec{r} = (4, 0, -1) + \mu(2, 0, 3)$. Let the point $P$ be $(43, \alpha, \beta)$. The line passing through $P$ with direction ratios $(3, -1, 0)$ is $\frac{x-43}{3} = \frac{y-\alpha}{-1} = \frac{z-\beta}{0} = \lambda$. Any point on this line is $P_1(43+3\lambda, \alpha-\lambda, \beta)$. Since $P_1$ lies on the line $\vec{r} = (4+2\mu, 0, -1+3\mu)$,we have: $43+3\lambda = 4+2\mu \Rightarrow 2\mu - 3\lambda = 39$ $\alpha-\lambda = 0 \Rightarrow \lambda = \alpha$ $\beta = -1+3\mu$ From $\lambda = \alpha$,we have $2\mu - 3\alpha = 39 \Rightarrow \mu = \frac{3\alpha+39}{2}$. Then $\beta = -1 + 3(\frac{3\alpha+39}{2}) = \frac{-2+9\alpha+117}{2} = \frac{9\alpha+115}{2}$. The distance $PP_1 = 13\sqrt{10}$,so $(PP_1)^2 = 1690$. $PP_1^2 = (3\lambda)^2 + (-\lambda)^2 + 0^2 = 10\lambda^2 = 1690 \Rightarrow \lambda^2 = 169 \Rightarrow \lambda = \pm 13$. If $\lambda = 13$,then $\alpha = 13$ and $\beta = \frac{9(13)+115}{2} = \frac{117+115}{2} = 116$ (not possible since $\beta If $\lambda = -13$,then $\alpha = -13$ and $\beta = \frac{9(-13)+115}{2} = \frac{-117+115}{2} = -1$. Thus,$\alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170$.

If the distance of the point $P(43, \alpha, \beta), \beta < 0$,from the line $\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}), \mu \in R$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$,then $\alpha^{2} + \beta^{2}$ is equal to . . . . . . .

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