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otez555 [7]
2 years ago
7

Find a decomposition of a=⟨−5,−1,1⟩ into a vector c parallel to b=⟨−6,0,6⟩ and a vector d perpendicular to b such that c+d=a.

Mathematics
1 answer:
dezoksy [38]2 years ago
7 0

The projection of vector A <em>parallel</em> to vector B is \langle -3, 0, 3\rangle and the projection of vector A <em>perpendicular</em> to vector B is \langle -2, -1, -2\rangle.

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 <em>parallel</em> to vector B is defined by this formula:

\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b (1)

Where \|\vec b\| is the norm of vector B.

And the projection of vector A <em>perpendicular</em> to vector B is:

\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b} (2)

If we know that a = \langle -5, -1, 1 \rangle and \vec b = \langle -6, 0, 6 \rangle, then the projections are now calculated:

\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

\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle

\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle

\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle

\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle

The projection of vector A <em>parallel</em> to vector B is \langle -3, 0, 3\rangle and the projection of vector A <em>perpendicular</em> to vector B is \langle -2, -1, -2\rangle.

We kindly invite to check this question on projection of vectors: brainly.com/question/24160729

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