Question

Find a vector of magnitude 3 and perpendicular to both the vectors b = 2i - 2j + k and c = 2i + 2j + 3k.

Answer 1

Answer:

$\vec r=2\hat i+\hat j-2\hat k\\\\or\\\\\vec r=-2\hat i-\hat j+2\hat k$

Step-by-step explanation:

To find a vector of magnitude 3 and perpendicular to both the vectors b = 2i - 2j + k and c = 2i + 2j + 3k.

let the vector is

$\vec r= x\hat i+y\hat j+z\hat k\\\\|\vec r|=3\\\\\sqrt{x^{2}+y^{2}+z^{2}} =3\\\\x^{2}+y^{2}+z^{2}=9 .......eq1$

Since vector r is perpendicular to both vector a and b

Thus dot product of vector r with b and c would be zero

2x-2y+z=0  ....eq2

2x+2y+3z=0....eq3

from equation 2 and 3

$\frac{x}{-6-2} =\frac{y}{2-6} =\frac{z}{4+4} \\\\\frac{x}{-8}=\frac{y}{-4}=\frac{z}{8}$

or

$\frac{x}{-2}=\frac{y}{-1} =\frac{z}{2} =k\\ \\x=-2k\\\\y=-k\\\\z=2k$

put these value of x,y,z in eq 1

$(-2k)^{2}+(-k)^{2}+(2k)^{2}=9\\\\4k^{2}+k^{2}+4k^{2}=9$

$9k^{2}=9\\\\k^{2}=1\\\\k=+1,-1$

So x=2,y=1,z=-2

or x=-2,y=-1,z=2

So vector is

$\vec r=2\hat i+\hat j-2\hat k\\\\or\\\\\vec r=-2\hat i-\hat j+2\hat k$

Answer 2

set by step explaination

Step-by-step explanation:

it will help full