Question

Let c be the unit vector coplanar with a=i - j + 2k and b= 2i - j + k such that c is perpendicular to a. If the projection of c along b is √11/k then k is

Answer 1

Answer:

a 6 and 35 and k=8 j+k and

Step-by-step explanation:

y 5 a3 k10 and they a1 a2 ~b1~b2

Answer 2

Answer:

6

Step-by-step explanation:

Given:

a=i - j + 2k

b= 2i - j + k

To find:

Value of k

Solution:

Vectors that are situated on the same surface of a three-dimensional plane are referred to as coplanar vectors. Each vector is perpendicular to the same plane. Any two random vectors in a plane that are coplanar are always simple to find. The location of coplanarity between two lines is in a three-dimensional vector space. When the scalar product of three vectors is zero, they are said to be coplanar.

A frequently discussed subject in three-dimensional geometry is coplanar lines. Three lines situated on the same plane are said to be coplanar when they satisfy the condition known as the coplanarity of three vectors in mathematical theory.

While we have utilised the straight lines as vector equations, a plane is a two-dimensional figure that extends into infinity in a three-dimensional region.

Let c= xa+yb

c=i(x+2y)+j(-x-y)+k(2x+y)

c.a=0

$x+2y+x+y+4x+2y=0$

$y=-\frac{6x}{5}$

$c=\frac{-7x}{5}i+\frac{x}{5}j+\frac{4x}{5} k\\ |c|=1\\=\frac{49x^{2} +x^{2} +16x^{2} }{25} \\=x^{2} =\frac{25}{66}$

c=±$\frac{5}{\sqrt{66} } ( \frac{-7}{5}i+ \frac{1}{5}j+ \frac{4}{5}k )$

The distance between the perpendiculars is,

$P=\frac{c.d}{|b|}=\frac{\sqrt{11} }{6}$

Thus the value of k is 6.

Define the following

1) coplanar vector

2) collinear vector

3) equal vector

4) negative vector

5) zero vector

6) unit vector

https://brainly.in/question/3526098

The value of 'p' for which the vectors 2i - j +k, 3i + pj +5k ( i, j, k are unit vectors) are

coplanar is

https://brainly.in/question/1192148

#SPJ2