Question

a vector normal to i+2j-k is​

Answer 1

Answer:

The cross product of two vectors, A⃗ ×B⃗ always gives a third vector that is perpendicular to both A⃗ and B⃗ . You can use this to get a vector that is perpendicular to your specified vector. Call the vector you specified as A⃗ .

A⃗ =(1,−2,1)

The trick is calculating a suitable B⃗ . It is important to ensure that B⃗ is not in the same direction as A⃗ or the cross product will go to zero. Anything that ensures this will work. One thing you can do is scramble the order of the components of A⃗ to get B⃗ . For example,

B⃗ =(2Az,−1.5Ax,4Ay)

Now you can calculate a third vector C⃗ as the cross product

C⃗ =A⃗ ×B⃗

Here you are guaranteed that C⃗ will be perpendicular to A⃗ .

So, following this recipe,

B⃗ =(2,−1.5,−8)

C⃗ =(1,−2,1)×(2,−1.5,−8)=(17.5,10,2.5)

Answer 2

Given,

$\[\overrightarrow F = \widehat i + 2\widehat j - \widehat k\]$

Solution,

Therefore, the normal vector

$\[\begin{array}{l} \Rightarrow {\widehat a_F} = \frac{{\widehat i + 2\widehat j - \widehat k}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2}} }}\\ \Rightarrow {\widehat a_F} = \frac{{\widehat i + 2\widehat j - \widehat k}}{{\sqrt {1 + 4 + 1} }}\\ \Rightarrow {\widehat a_F} = \frac{{\widehat i + 2\widehat j - \widehat k}}{{\sqrt 6 }}\end{array}\]$

Hence, the normal to the given vector is $\[\frac{{\widehat i + 2\widehat j - \widehat k}}{{\sqrt 6 }}.\]$