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The sine of the angle between the straight line \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} and

The plane 2x - 2y + z = 5 is:
A. 10/6√5
B. 4/5√2
C. 2√3/5
D. √2/10

Answers (1)

The equation of the line is given as


The direction vector of this line can be represented as \vec{b}=3\hat{i}+4 \hat{j}+5\hat{k}

Also given is the equation of the plane 2x - 2y + z = 5

The normal to this plane is,\vec{n}=2\hat{i}-2 \hat{j}+\hat{k}

We also know that the angle \phi between the line with the direction        

   vector b and the plane with the normal vector n is,

\sin \phi =\left | \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |} \right |

\Rightarrow \sin \phi =\left |\frac{3(2)+4(-2)+5(1)}{\sqrt{3^{2}+4^{2}+5^{2}}\sqrt{2^{2}+(-2)^{2}+1^{2}}} \right |

\Rightarrow \sin \phi =\left |\frac{3}{3\sqrt{50}} \right |

=\frac{1}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}


(Option D)

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