Problem 3. (1 point) Find the orthogonal projection of vecv...

Problem 3. (1 point) Find the orthogonal projection of vec{v} = egin{bmatrix} 5 \ -3 \ -2 \ 0 end{bmatrix} onto the subspace W of mathbb{R}^4 spanned by egin{bmatrix} -1 \ -1 \ -1 \ 1 end{bmatrix} and egin{bmatrix} 0 \ 1 \ -1 \ 0 end{bmatrix}. proj_W(vec{v}) = egin{bmatrix} Box \ Box \ Box \ Box end{bmatrix}

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00:01 So, to find the orthogonal projection of v under the subspace spanned by these two vectors u and u2, the first thing we need to check is whether u1, u2, or orthogonal, when do you say two vectors orthogonal when the dot product is 0.

00:19 So let's find the dot product.

00:21 Negative 1 into 0 is 0, so negative 1 into 0 plus negative 1 into 1, negative 1, it is negative 1, is negative 1, it is negative 1, positive 1, and 1.

00:35 To 0 0.

00:36 So basically you'll get negative 1 plus 1 yes they're also.

00:40 So the projection of w projection of v on the w is basically given by v .2 divided by mod u1 squared into u1 vector plus v dot u2 divided by mod u2 square into u2 vector.

01:06 So this is the vector projection of v on to w where u and you two are orthogonal vectors.

01:18 Now what is v.

01:19 U1? so take dot product of v and u1.

01:22 It will be negative 5 plus 3, negative 5 plus 3 plus 2 plus 0.

01:31 So it is basically 0.

01:34 So even v and u .1 are orthog.

01:36 So that means this is 0.

01:37 The first term is completely 0.

01:39 So it's a 0 vector.

01:41 Now let's do v .2.

01:42 2, v .u2 will be dot product of v and u2, 0 into 5, 1 into negative 3, negative 2, negative 2, negative 2, and 0 and 0 0, 0.

01:55 So basically it is negative 1.

01:57 So you get negative 1 divided by modulus of u2 square...

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