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Can someone please help me out?
40POINTS 2 QUESTIONS ONLY!

Can someone please help me out? 40POINTS 2 QUESTIONS ONLY!-example-1
Can someone please help me out? 40POINTS 2 QUESTIONS ONLY!-example-1
Can someone please help me out? 40POINTS 2 QUESTIONS ONLY!-example-2
User RobertR
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1 Answer

5 votes

Answer:

Q1: θ ≈ 25°

Q2: First option, parallel; type "PARALLEL"

Explanation:

To find the cosine of the angle between two planes, we can use the following formula:


\rightarrow \theta = \cos^(-1)\Big((a_xb_x+a_yb_y+z_xz_y)/(\big|\big|\vec a \big|\big|\big|\big|\vec b \big|\big|)\Big); \ 0\textdegree \leq \theta \leq 180 \textdegree

Where vectors a and b are the vectors normal to the given planes.


\hrulefill

Q1: Find the cosine of the angle between the planes x + y + z = 0 and x + 3y + 4z = 3.

Step 1: Find the normal vectors of the planes:

For the given plane, x + y + z = 0:

The coefficients of x, y, and z represent the normal vector, so the normal vector is <1, 1, 1>.

For the plane x + 3y + 4z = 3:

The coefficients of x, y, and z represent the normal vector, so the normal vector is <1, 3, 4>.

Step 2: Calculate the dot product of the normal vectors:

⇒ Dot product = (1 · 1) + (1 · 3) + (1 · 4) = 1 + 3 + 4 = 8

Step 3: Calculate the magnitudes of the normal vectors:

⇒ Magnitude of the normal vector of the first plane = √((1)² + (1)² + (1)²) = √3.

⇒ Magnitude of the normal vector of the second plane = √((1)² + (3)² + (4)²) = √26.

Step 4: Find the cosine of the angle between the planes:


\Longrightarrow \theta = \cos^(-1)\Big((8)/(√(3) \cdot√(26) ) \Big)\\\\\\\\\Longrightarrow \theta = \cos^(-1)\Big((8)/(√(78) ) \Big)\\\\\\\\\therefore \boxed{\theta \approx 25 \textdegree}
\hrulefill

Q2: Determine whether the planes 15x - 6y + 12z = 2 and 4y = 10x + 8z are parallel, perpendicular, or neither.

Step 1: Write both planes in the form Ax + By + Cz = D:

The plane, 15x - 6y + 12z = 2, is already written in this form.

The plane, 4y = 10x + 8z, is rewritten as 10x - 4y + 8z = 0.

Step 2: Find the normal vectors of the planes:

For the plane 15x - 6y + 12z = 2:

The coefficients of x, y, and z represent the normal vector, so the normal vector is <15, -6, 12>.

For the plane 10x - 4y + 8z = 0:

The coefficients of x, y, and z represent the normal vector, so the normal vector is <10, -4, 8>.

Step 3: Check if the normal vectors are parallel or perpendicular:

  • Two vectors are parallel if their cross product is the zero vector.
  • Two vectors are perpendicular if their dot product is zero.

Taking the cross product:


\Longrightarrow \left|\begin{array}{ccc}\hat i &amp; \hat j &amp; \hat k\\15&amp; -6&amp;12\\10&amp;-4&amp;8\end{array}\right| = \left|\begin{array}{cc}-6&amp;12\\-4&amp;8\end{array}\right| \hat i \ - \left|\begin{array}{cc}15&amp;12\\10&amp;8\end{array}\right| \hat j \ + \left|\begin{array}{cc}15&amp;-6\\10&amp;-4\end{array}\right| \hat k\\\\\\\\\Longrightarrow [(-6)(8)-(12)(-4)] \hat i - [(15)(8)-(12)(10)] \hat j + [(15)(-4)-(-6)(10)] \hat k


\Longrightarrow [-48+48] \hat i - [120-120] \hat j + [-60+60] \hat k\\\\\\\\\Longrightarrow0 \hat i - 0\hat j + 0\hat k \therefore \text{parallel}

Thus, these planes are parallel. The second box wants you to input PARALLEL.


\hrulefill

Additional information:

Normal Vector: The normal vector of a plane is a vector that is perpendicular to the plane. For a plane in the form Ax + By + Cz = D, the normal vector is <A, B, C>.

Cross Product: The cross product of two vectors in three-dimensional space results in a vector that is perpendicular to both of the original vectors. If the cross product of two vectors is the zero vector, it means the vectors are parallel.


\boxed{\left\begin{array}{ccc}\text{\underline{Cross Product:}}\\\text{Given two vectors,} \ \vec a \ \text{and} \ \vec b:\\\\\vec a * \vec b = \left|\begin{array}{ccc}\hat i &amp;\hat j &amp;\hat k\\a_x&amp;a_y&amp;a_z\\b_x&amp;b_y&amp;b_z\end{array}\right| = \left|\begin{array}{cc}a_y&amp;a_z\\b_y&amp;b_z\end{array}\right| \hat i \ - \left|\begin{array}{cc}a_x&amp;a_z\\b_x&amp;b_z\end{array}\right| \hat j \ + \left|\begin{array}{cc}a_x&amp;a_y\\b_x&amp;b_y\end{array}\right| \hat k\ \end{array}\right }

Dot Product: The dot product of two vectors is a scalar value that represents the cosine of the angle between the vectors when they are normalized (divided by their magnitudes). If the dot product is zero, the vectors are perpendicular.


\boxed{\left\begin{array}{ccc}\text{\underline{Dot Product:}}\\\text{Given two vectors,} \ \vec a \ \text{and} \ \vec b:\\\\ \vec a \cdot\vec b = a_xb_x+a_yb_y+a_zb_z \end{array}\right }

User Dhwani
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