Question

Find a unit vector orthogonal to (4, 2, 3) in R3​

Answer 1

Answer:

9

Step-by-step explanation:

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Answer 2

Given a vector $(4,2,3)$, Find a unit vector orthogonal to it.

Explanation:

• R3 here means that the three dimensional space is being considered here that has three axes, x, y, z.
• Now a point (x, y, z) in a three dimensional space can be represented as a vector 'v' by, $\vec v=x\hat i+y\hat j+z\hat k$
• Hence the given point can be vector represented by, $\vec v=4\hat i+2\hat j+3\hat k$
• Let the required unit vector be denoted 'r' and be given by, $\vec r=x\hat i+y\hat j+z\hat k$  
• Now since its a unit vector we have, $|\vec r|=\sqrt{x^2+y^2+z^2}=1$   ---(a)
• For the vectors 'v' and 'r' to be orthogonal their dot product should be equal to zero,
• $\vec v.\vec r=|\vec v||\vec r|cos(90)=0\\->(4\hat i+2\hat j+3\hat k)(x\hat i+y\hat j+z\hat k)=0\\ -> \hat i . \hat i=\hat j . \hat j=\hat k . \hat k=1,\ \ \hat i . \hat j=\hat j . \hat k=\hat k . \hat i=0\\ ->4x+2y+3z=0$   ----(b)
• Solving (a) and (b) we get,   $x=\frac{1}{\sqrt{6} } ,\ \ y=\frac{1}{\sqrt{6} } ,\ \ z=-\sqrt{\frac{2}{3} }$  
• Hence the unit vector orthogonal to the given vector is $\vec r= \frac{1}{\sqrt{6} } \hat i+ \frac{1}{\sqrt{6} } \hat j+ -\sqrt{\frac{2}{3 } }\hat k$