The distance between the point (3, 4, 5) and the point where the line x - 3 1 = y - 4 2 = z - 5 2 meets the plane x + y + z = 17 , is:

Question

The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane $x + y + z = 17$ , is:

Smart Achievers · Accepted Answer

(1)

Ans. 3

Explanation:

$\frac{x_{0} - 3}{1} = \frac{y_{0} - 4}{2} = \frac{z_{0} - 5}{2} = k$

$x_{0} = k + 3, y_{0} = 2 k + 4, z_{0} = 2 k + 5$

$We have, x_{0} + y_{0} + z_{0} = 17$

$5 k + 12 = 17$

$k = 1$

$x_{0} = 4, y_{0} = 6, z_{0} = 7$

$point = (4, 6, 7)$

$Distance between (3, 4, 5) and (4, 6, 7)$

$= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} + {(z_{1} - z_{2})}^{2}}$

$= \sqrt{1 + 4 + 4}$

$= 3$

Solution

Q.
The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane $x + y + z = 17$ , is:

1 3

2 2

3 1

4 0

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Solution

Q. The distance between the point (3, 4, 5) and the point where the line x-31=y-42=z-52 meets the plane x+y+z=17, is:

1 3

2 2

3 1

4 0

Ans. (1) Ans. 3 Explanation: x0-31=y0-42=z0-52=k x0=k+3, y0=2k+4, z0=2k+5 We have, x0+y0+z0=17 5k+12=17 k=1 x0=4, y0=6, z0=7 point=(4,6,7) Distance between (3,4,5) and (4,6,7) =(x1-x2)2+(y1-y2)2+(z1-z2)2 =1+4+4 =3

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Q.
The distance between the point (3, 4, 5) and the point where the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ meets the plane $x + y + z = 17$ , is: