How do you compute the dot product to find the magnitude of #u=<-5,12>#?

Answer 1

#color(blue)(13)#

You cannot compute the dot product with only 1 vector. You do not need the dot product to find the magnitude of the given vector anyway.

The magnitude of a position vector #bba=[x_1,y_1]# is given by:

#||bba||=sqrt((x_1)^2+(y_1)^2#

Hence:

#bbu=[-5,12]#

#||bbu||=sqrt((-5)^2+(12)^2)=sqrt(169)=13#

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Answer 2

Sadie Ackerman

To find the magnitude of a vector \( \mathbf{u} = <-5, 12> \), you compute the dot product of the vector with itself and then take the square root of the result: \[ |\mathbf{u}| = \sqrt{\mathbf{u} \cdot \mathbf{u}} \] In this case, the dot product is: \[ \mathbf{u} \cdot \mathbf{u} = (-5) \cdot (-5) + 12 \cdot 12 \] \[ \mathbf{u} \cdot \mathbf{u} = 25 + 144 \] \[ \mathbf{u} \cdot \mathbf{u} = 169 \] Therefore, the magnitude of vector \( \mathbf{u} \) is: \[ |\mathbf{u}| = \sqrt{169} \] \[ |\mathbf{u}| = 13 \]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you compute the dot product to find the magnitude of #u=&lt;-5,12&gt;#?

How do you compute the dot product to find the magnitude of #u=<-5,12>#?