What is the projection of #<-2,2,5 ># onto #<1,6,9 >#?

Answer 1

The answer is #=55/118〈1,6,9〉#

The projection of #vecb# onto #veca# is
#=(veca.vecb)/(∥veca∥^2)veca#
The dot product is #=〈-2,2,5〉.〈1,6,9〉#
#=(-2+12+45)#
#=55#
The modulus of #veca#
#=∥〈1,6,9〉∥#
#=sqrt(1+36+81)=sqrt118#

Consequently,

The forecast is

#=55/118〈1,6,9〉#
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Answer 2
To find the projection of vector \( \vec{v} \) onto vector \( \vec{u} \), you can use the formula: \( \text{proj}_{\vec{u}}(\vec{v}) = \frac{\vec{v} \cdot \vec{u}}{\lVert \vec{u} \rVert^2} \vec{u} \). Given \( \vec{v} = <-2, 2, 5> \) and \( \vec{u} = <1, 6, 9> \), the projection of \( \vec{v} \) onto \( \vec{u} \) is \( \text{proj}_{\vec{u}}(\vec{v}) = \frac{<-2, 2, 5> \cdot <1, 6, 9>}{\lVert <1, 6, 9> \rVert^2} <1, 6, 9> \). Calculating the dot product of \( \vec{v} \) and \( \vec{u} \) yields \( <-2, 2, 5> \cdot <1, 6, 9> = -2(1) + 2(6) + 5(9) = 52 \). The magnitude of \( \vec{u} \) is \( \lVert <1, 6, 9> \rVert = \sqrt{1^2 + 6^2 + 9^2} = \sqrt{1 + 36 + 81} = \sqrt{118} \). Thus, the projection of \( \vec{v} \) onto \( \vec{u} \) is \( \frac{52}{118} <1, 6, 9> = <\frac{26}{59}, \frac{156}{59}, \frac{234}{59}> \).
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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.

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.

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.

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.

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