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Mumz [18]
3 years ago
7

Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5. Find a vector v¯¯¯ that is

perpendicular to p.
Mathematics
1 answer:
Lana71 [14]3 years ago
7 0

Let P be the plane that intersects

  • x-axis at point (-5,0,0);
  • y-axis at point (0,-2,0);
  • z-axis at point (0,0,5).

Write the equation of the plane P:

\left|\begin{array}{ccc}x-(-5)&y-0&z-0\\0-(-5)&-2-0&0-0\\0-(-5)&0-0&5-0\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x+5&y&z\\5&-2&0\\5&0&5\end{array}\right|=0.

Then

-10(x+5)+10z-25y=0,\\ \\2(x+5)-2z+5y=0,\\ \\2x+5y-2z=-10.

The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus

\vec{v}=(2,5,-2)\perp P.

Answer: \vec{v}=(2,5,-2).

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With bases V1, V2 , V3 andw1, w2 , W3, suppose T(v1) = w2 and T(v2) = T(v3) = w1 + w3 . T is a linear transformation. Find the m
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Answer:

See picture and explanation below.

Step-by-step explanation:

With this information, the matrix A that you can find is the transformation matrix of T. The matrix A is useful because T(x)=Av for all v in the domain of T.

A is defined as T=([T(v_1)] [T(v_2] [T(v_3)])\text{ where }[T(v_i)] denotes the vector of coordinates of T(v_i) respect to the basis (we can apply this definition because forms a basis for the domain of T).

The vector of coordinates can be computed in the following way: if T(v_i)=a_1w_1+a_2w_2+a_3w_3 then [T(v_i)]=(a_1,a_2,a_3)^t.

Note that we have all the required information: T(v_1)=0w_1+1\cdot w_2+0w_3 then [T(v_1)]=(0,1,0)^t

T(v_2)=T(v_3)=1\cdot w_1+0w_2+0w_3 hence [T(v_2)]=[T(v_3)]=(1,0,0)^t

The matrix A is on the picture attached, with the multiplication A(1,1,1).

Finally, to obtain the output required at the end, use the properties of a linear transformation and the outputs given:

T(v_1+v_2+v_3)=T(v_1)+T(v_2)+T(v_3)=w_2+w_1+w_1=w_2+2w_1  

In this last case, we can either use the linearity of T or multiply by A.

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