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M1 vectors question

20170222_154927.jpg

How do you do 5 (a) ?

Btw the cake eraser is just there for decoration.

Thanks 😊
That question looks delicious.
Original post by LeCroissant
20170222_154927.jpg

How do you do 5 (a) ?

Btw the cake eraser is just there for decoration.

Thanks 😊


Find lambda such that the j vector is 0
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Zacken
22
Original post by LeCroissant
20170222_154927.jpg

How do you do 5 (a) ?

Btw the cake eraser is just there for decoration.

Thanks 😊


First off, for the vector space R2\mathbb{R}^2, we can represent a vector a=a1i+a2j\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} where i\mathbf{i} and j\mathbf{j} are orthogonal (perpendicular) vectors, usually taken to be unit vectors in the directions of the x and y-axis.

Anyway. We say two vectors a=a1i+a2j\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} and b=b1i+b2j\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} are equal if their components are equal, i,e: a=b  ⟺  a1=b1 and a2=b2\mathbf{a} = \mathbf{b} \iff a_1 = b_1 \, \text{and}\, a_2 = b_2.

Secondly, we say two vectors are parallel if there exists a real, non-zero scalar kk such that a=kb\mathbf{a} = k\mathbf{b}.

In your case, for a+λb=(2+3λ)i+(5−λ)j\mathbf{a} + \lambda \mathbf{b} = (2+3\lambda)\mathbf{i} + (5 - \lambda) \mathbf{j} to be parallel to 1i+0j1\mathbf{i + 0\mathbf{j}} we want to determine the existence of a kk such that (2+3λ)i+(5−λ)j=ki+0kj(2+3\lambda)\mathbf{i} + (5 - \lambda) \mathbf{j}= k\mathbf{i} + 0k \mathbf{j}.

This gives us 5−λ=05-\lambda = 0 and 2+3λ=k2 + 3\lambda = k. This can be solved simultaneously to give you λ\lambda and kk.

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