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  • Help me solve this The distance of the point (1, −5, 9) from the plane x − y + z = 5 measured along theline x = y = z is :

The distance of the point (1, −5, 9) from the plane x − y + z = 5 measured along the

line x = y = z is :

  • Option 1)

    3\sqrt{10}

  • Option 2)

    10\sqrt{3}

  • Option 3)

    \frac{10}{\sqrt{3}}

  • Option 4)

    \frac{20}{3}

 

Answers (1)

best_answer

As we learnt in 

Distance formula -

The distance between two points A(x_{1},y_{1},z_{1})\, and \, B(x_{2},y_{2},z_{2}) is =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+{\left ( y_{2}-y_{1} \right )^{2}}+{\left ( z_{2}-z_{1} \right )^{2}}}

 

- wherein

\vec{OA}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}

\vec{OB}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

\underset{AB}{\rightarrow}    = \underset{OB}{\rightarrow} - \underset{AB}{\rightarrow}

\underset{AB}{\rightarrow}  = \left ( x_{2}-x_{1} \right )\hat{i}+\left ( y_{2}-y_{1} \right )\hat{j}+\left ( z_{2}-z_{1} \right )\hat{k}

AB= \left | \underset{AB}{\rightarrow} \right |

AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

 

 

 

 

Cartesian eqution of a line -

The equation of a line passing through two points A\left ( x_{0},y_{0},z_{0} \right )and parallel to vector having direction ratios as \left ( a,b,c \right )is given by

\frac{x-x_{0}}{a}= \frac{y-y_{0}}{b}= \frac{z-z_{0}}{c}

The equation of a line passing through two points A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right ) is given by

\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

- wherein

 

 Coordinates of point A

Line AP is x-1=y+5=z-9=k

Thus, x=k+1, y=k-5, z=k+9

x-y+z=5

means k+1-k-5+k+9=5

k=-10

Point A is (-9,-15,-1)

AP is \sqrt{10^{2}+10^{2}+10^{2}}=10\sqrt{3}


Option 1)

3\sqrt{10}

This option is incorrect

Option 2)

10\sqrt{3}

This option is correct

Option 3)

\frac{10}{\sqrt{3}}

This option is incorrect

Option 4)

\frac{20}{3}

This option is incorrect

Posted by

Aadil

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