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  • A vector  in the first octant is inclined to the <img alt

A vector \overrightarrow{\mathrm{v}} in the first octant is inclined to the x- axis at 60^{\circ} , to the y-axis at 45 and to the z-axis at an acute angle. If a plane passing through the points (\sqrt{2},-1,1) and (a, b, c), is normal to \vec{v},then

Option: 1

\sqrt{2} a+b+c=1


Option: 2

a+\sqrt{2} b+c=1


Option: 3

a+b+\sqrt{2} c=1


Option: 4

\sqrt{2} a-b+c=1


Answers (1)

\cos \alpha=\cos 60                            \cos \beta=\cos 45
\ell=\frac{1}{2}                                                \mathrm{m}=\frac{1}{\sqrt{2}}
\mathrm{l}^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=1

\Rightarrow \frac{1}{4}+\frac{1}{2}+\mathrm{n}^{2}=1           n^{2}=1-\frac{3}{4}=\frac{1}{4}
                                                       \mathrm{n}=\frac{1}{2}

Direction of \overrightarrow{\mathrm{v}} is \frac{1}{2} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{2} \hat{k}

Equation of plane through (\sqrt{2},-1,1) &  Normla to \overrightarrow{\mathrm{v}} is 
\frac{1}{2}(\mathrm{x}-\sqrt{2})+\frac{1}{\sqrt{2}}(\mathrm{y}+1)+\frac{1}{2}(\mathrm{z}-1)=0

It passes through (a,b,c)

(a-\sqrt{2})+\sqrt{2}(b+1)+(c-1)=0
\Rightarrow \mathrm{a}+\sqrt{2} \mathrm{~b}+\mathrm{c}=\sqrt{2}-\sqrt{2}+1
\Rightarrow a+\sqrt{2} b+c=1

Posted by

Sumit Saini

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