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The projection of vector A parallel to vector B is [tex]\langle -3, 0, 3\rangle[/tex] and the projection of vector A perpendicular to vector B is [tex]\langle -2, -1, -2\rangle[/tex].
In this question, we need to determine all projections of a vector with respect to another vector. In this case, the projection of vector A parallel to vector B is defined by this formula:
[tex]\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b[/tex] (1)
Where [tex]\|\vec b\|[/tex] is the norm of vector B.
And the projection of vector A perpendicular to vector B is:
[tex]\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b}[/tex] (2)
If we know that [tex]a = \langle -5, -1, 1 \rangle[/tex] and [tex]\vec b = \langle -6, 0, 6 \rangle[/tex], then the projections are now calculated:
[tex]\vec a_{\parallel, \vec b} = \frac{(-5)\cdot (-6)+(-1)\cdot (0)+(1)\cdot (6)}{(-6)^{2}+0^{2}+6^{2}} \cdot \langle -6, 0, 6 \rangle[/tex]
[tex]\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle[/tex]
[tex]\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle[/tex]
[tex]\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle[/tex]
[tex]\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle[/tex]
The projection of vector A parallel to vector B is [tex]\langle -3, 0, 3\rangle[/tex] and the projection of vector A perpendicular to vector B is [tex]\langle -2, -1, -2\rangle[/tex].
We kindly invite to check this question on projection of vectors: https://brainly.com/question/24160729
