Let I be the plane passing through the points R=(2,-1,1), S=(1,2,-1) and T=(-3,1,2). Let U be the plane perpendicular to I and passing through the points S and (0, 1, 1). The cosine of the angle between U and I is given by: \( {1}{ {2}}\) \(0\) \(1\) None of the given options \(-1\)See answer - CODE PELAJAR

Let I be the plane passing through the points R 2 11 S 12 1 and T 312 Let U be the plane perpendicular to I and passing through the points S and 0 1 1 The cosine of the angle between U and I is given…

Key Concept

Planes’ Angle

UndergraduateMathLinear Algebra

The cosine of the angle between two planes is the absolute value of the dot product of their normal vectors, divided by the product of their magnitudes.

Basic Answer

Ideas for solving the problem：

This problem can be solved by determining the normal vectors of the planes $I$ and $U$, and then using the dot product formula to find the cosine of the angle between these planes.

Calculation step：

Step 1: Find the normal vector of plane $I$
To find the normal vector of plane $I$, we first find two vectors in the plane by subtracting the coordinates of the points $R$, $S$, and $T$.

Vector $RS$:
$$RS=S-R=(1-2,2-(-1),-1-1)=(-1,3,-2)$$

Vector $RT$:
$$RT=T-R=(-3-2,1-(-1),2-1)=(-5,2,1)$$

The normal vector $\overrightarrow{n_I}$ of plane $I$ is the cross product of vectors $RS$ and $RT$:
$$\overrightarrow{n_I}=RS\times RT=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -1& 3& -2\\ -5& 2& 1\end{array}\right|=\mathbf{i}(3\cdot 1-(-2)\cdot 2)-\mathbf{j}((-1)\cdot 1-(-2)\cdot (-5))+\mathbf{k}((-1)\cdot 2-3\cdot (-5))$$
$$=\mathbf{i}(3+4)-\mathbf{j}(-1-10)+\mathbf{k}(-2+15)=\mathbf{i}(7)-\mathbf{j}(-11)+\mathbf{k}(13)=(7,11,13)$$

Step 2: Find the normal vector of plane $U$
Plane $U$ is perpendicular to plane $I$ and passes through points $S$ and $(0,1,1)$. The normal vector $\overrightarrow{n_U}$ of plane $U$ is the same as the normal vector $\overrightarrow{n_I}$ of plane $I$ because $U$ is perpendicular to $I$.

$$\overrightarrow{n_U}=(7,11,13)$$

Step 3: Calculate the cosine of the angle between the planes
The cosine of the angle between the planes is given by the dot product of their normal vectors divided by the product of their magnitudes.

$$\cos (\theta )=\frac{\overrightarrow{n_I}\cdot \overrightarrow{n_U}}{|\overrightarrow{n_I}||\overrightarrow{n_U}|}$$

Since $\overrightarrow{n_I}=\overrightarrow{n_U}$, the dot product $\overrightarrow{n_I}\cdot \overrightarrow{n_U}=\overrightarrow{n_I}\cdot \overrightarrow{n_I}$:
$$\overrightarrow{n_I}\cdot \overrightarrow{n_I}=7^2+11^2+13^2=49+121+169=339$$

The magnitude of $\overrightarrow{n_I}$:
$$|\overrightarrow{n_I}|=\sqrt{7^2+11^2+13^2}=\sqrt{339}$$

Thus,
$$\cos (\theta )=\frac{339}{\sqrt{339}\cdot \sqrt{339}}=\frac{339}{339}=1$$

Final Answer:

The cosine of the angle between planes $U$ and $I$ is $1$.

Highlights:

• The normal vectors of perpendicular planes are the same.
• The cosine of the angle between two vectors is calculated using the dot product formula.

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Answered Nov 21 at 04:24 by StudyX AI with Basic Model

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