Math, asked by arijit36, 1 year ago

let ' n ' vector be a vector of magnitude 2√3 such that it makes equal acute angles with the co-ordinate Axes find the vector and cartesian forms of the equation of a plane passing through ( 1, - 1, 2 ) and normal to 'n' vector.

Answers

Answered by CarlynBronk
13

Magnitude of vector 'n'= 2\sqrt{3}

It is given that vector makes equal angles with coordinate axes.If l, m, and n are direction cosines of a line then

l=Cos A, m =Cos B, n=Cos C

As it is given that the vector n makes equal angle with coordinate axes,

So, A=B=C

As we know

l²+m²+n²=1

Substituting the value of l, m and n, we get

Cos²A+Cos²A+Cos²A=1

3 Cos²A=1

→Cos A=\pm\frac{1}{\sqrt{3}}

So the Vector n is =\pm2\sqrt{3} [\frac{i}{\sqrt{3}}+ \frac{j}{\sqrt{3}}+ \frac{k}{\sqrt{3}} ]

Direction ratios of normal is \pm(2,2,2)

Equation of plane passing through (1,-1,2) and normal to vector n having direction cosines \pm(2,2,2) is given by

→2(x-1) +2(y+1)+2(z-2)=0

→x-1+y+1+z-2=0

→x+y+z-2=0

x+y+z=2

Answered by IbadatCheema
8

Answer:

Cartesian form- x+y+z=2

Vector form- (i^+j^+k^).r =2

Attachments:
Similar questions