What is the unit vector that is orthogonal to the plane containing # (-i + j + k) # and # (i -2j + 3k) #?

Question

What is the unit vector that is orthogonal to the plane containing #  (-i + j + k) # and #  (i -2j + 3k)        #?

Lily Adams · Accepted Answer

The unit vector is #=<5/sqrt42,4/sqrt42,1/sqrt42>#

We perform a cross product to determine the vector that is perpendicular to the other two vectors.

Let #veca=<-1,1,1>#

#vecb=<1,-2,3>#

#vecc=|(hati,hatj,hatk),(-1,1,1),(1,-2,3)|#

#=hati|(1,1),(-2,3)|-hatj|(-1,1),(1,3)|+hatk|(-1,1),(1,-2)|#

#=hati(5)-hatj(-4)+hatk(1)#

#=<5,4,1>#

Confirmation

#veca.vecc=<-1,1,1>.<5,4,1>=-5+4+1=0#

#vecb.vecc=<1,-2,3>.<5,4,1>=5-8+3=0#

The modulus of #vecc=||vecc||=||<5,4,1>||=sqrt(25+16+1)=sqrt42#

The unit vector # = vecc /(||vecc||)#

#=1/sqrt42<5,4,1>#

Carson Alexander · Answer

undefined The unit vector orthogonal to the plane containing (-i + j + k) and (i - 2j + 3k) is (3i - 3j + 5k)/sqrt(35).

Amelia Adams · Answer

undefined To find the unit vector that is orthogonal (perpendicular) to the plane containing the vectors $ (-i + j + k) $ and $ (i - 2j + 3k) $:

1. **Find the Normal Vector:**
   - Calculate the cross product of the two given vectors to find a vector normal (orthogonal) to the plane they span.

2. **Cross Product:**
   - Let $ \mathbf{v}_1 = (-1, 1, 1) $ and $ \mathbf{v}_2 = (1, -2, 3) $ represent the given vectors.
   - Calculate the cross product: $ \mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 $.

3. **Normalize the Vector:**
   - To obtain a unit vector, divide the result by its magnitude.

4. **Magnitude of a Vector:**
   - Compute the magnitude of the cross product: $ |\mathbf{n}| = \sqrt{n_1^2 + n_2^2 + n_3^2} $.

5. **Unit Vector:**
   - Divide each component of $ \mathbf{n} $ by its magnitude: $ \frac{\mathbf{n}}{|\mathbf{n}|} $.

6. **Final Step:**
   - Express the resulting vector in terms of its components.

This unit vector will be orthogonal to the plane containing the given vectors $ (-i + j + k) $ and $ (i - 2j + 3k) $.