"Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at a distance of 9 units from the point $P$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :"

Question

Accepted Answer

"

Line through $P Q$

$\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4}$

Any point on $P Q$. be $R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)$

$P R=9$ unit

$(P R)^2=81$

$(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81$

$16 \lambda^2+4 \lambda^2+16 \lambda^2=81$

$36 \lambda^2=81$

$\lambda= \pm \frac{9}{6}= \pm \frac{3}{2}$

$\therefore R$ can be $(7,-5,9)$ or $(-5,1,-3)$

Distance from origin for both points be $\sqrt{49+25+81}$ and $\sqrt{25+1+9}=\sqrt{35}$

$\therefore$ Distance of $(7,-5,9)$ is farthest from origin

$\therefore(\alpha, \beta, \gamma)=(7,-5,9)$

Now $7^2+(-5)^2+9^2=155$

"

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at a distance of 9 units from the point $P$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Solution

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Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at a distance of 9 units from the point $P$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Solution

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