30) All the following points lie on the same side of the plane $$ \stackrel{\rightharpoonup}{r} \cdot(3 \hat{i}-5 \hat{j}+4 \hat{k})=5 $$ Except. $$ \qquad $$ a. $$ (1,2,1) $$ b- $$ (0,-1,2) $$ c. $$ (1,2,4) $$ d. $$ (2,-3,1) $$ 31) IF the intersecting Line of the two planes $$ 3 x-y+2 z+6=0, x+4 y-z+c=0 $$ Intersects $$ Z $$-axis, Then a. 3 b. -3 c. $$ \frac{1}{3} $$ d. $$ \frac{-1}{3} $$ 32) The measure of the angle between the Straight Line : $$ \frac{x-1}{2}=\frac{y+3}{-1} $$ and The plane $$ x+y+4=0 $$ is $$ \qquad $$ a. 0 b. 30 c- 60 d. 45 33) The Equation of the Line of intersection of the poplanes: $$ \begin{array}{l} x+y+z-6=0,2 x+3 y+4 z+5=0 \text { is } \ \text { a- } x=\frac{y-29}{2}=z+32 \quad \text { c. } x=\frac{-y+29}{2}=z+2 \cdot x=\frac{-y+29}{2}=z \end{array} $$

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UpStudy AI · Accepted Answer

Let's solve each of the problems step by step.

### Problem 30
We need to determine which point does not satisfy the plane equation:

$$
\stackrel{\rightharpoonup}{r} \cdot (3 \hat{i} - 5 \hat{j} + 4 \hat{k}) = 5
$$

This can be rewritten as:

$$
3x - 5y + 4z = 5
$$

We will evaluate each point to see if it satisfies this equation.

1. **Point (1, 2, 1)**:
   $$
   3(1) - 5(2) + 4(1) = 3 - 10 + 4 = -3 \quad (\text{not } 5)
   $$

2. **Point (0, -1, 2)**:
   $$
   3(0) - 5(-1) + 4(2) = 0 + 5 + 8 = 13 \quad (\text{not } 5)
   $$

3. **Point (1, 2, 4)**:
   $$
   3(1) - 5(2) + 4(4) = 3 - 10 + 16 = 9 \quad (\text{not } 5)
   $$

4. **Point (2, -3, 1)**:
   $$
   3(2) - 5(-3) + 4(1) = 6 + 15 + 4 = 25 \quad (\text{not } 5)
   $$

None of the points satisfy the equation, but we need to find the one that lies on the opposite side of the plane. The correct answer is the point that gives a different sign when substituted into the equation.

### Problem 31
We need to find the value of $$ c $$ such that the line of intersection of the two planes intersects the Z-axis.

The Z-axis is defined by $$ x = 0 $$ and $$ y = 0 $$. We will substitute these values into the equations of the planes:

1. For the first plane:
   $$
   3(0) - 0 + 2z + 6 = 0 \implies 2z + 6 = 0 \implies z = -3
   $$

2. For the second plane:
   $$
   0 + 4(0) - z + c = 0 \implies -z + c = 0 \implies z = c
   $$

Setting $$ z = -3 $$ from the first plane equal to $$ z = c $$ from the second plane gives us:

$$
c = -3
$$

### Problem 32
To find the angle between the line and the plane, we first need the direction vector of the line and the normal vector of the plane.

The line is given by:

$$
\frac{x-1}{2} = \frac{y+3}{-1} \implies \text{Direction vector } \mathbf{d} = (2, -1, 0)
$$

The plane is given by:

$$
x + y + 4 = 0 \implies \text{Normal vector } \mathbf{n} = (1, 1, 0)
$$

The angle $$ \theta $$ between the line and the plane can be found using the formula:

$$
\sin(\theta) = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}
$$

Calculating the dot product:

$$
\mathbf{d} \cdot \mathbf{n} = 2(1) + (-1)(1) + 0(0) = 2 - 1 = 1
$$

Calculating the magnitudes:

$$
|\mathbf{d}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5}
$$
$$
|\mathbf{n}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}
$$

Now substituting into the sine formula:

$$
\sin(\theta) = \frac{|1|}{\sqrt{5} \cdot \sqrt{2}} = \frac{1}{\sqrt{10}}
$$

To find the angle, we can use the inverse sine function. However, we can also find the angle between the line and the normal vector of the plane, which is complementary to the angle we want.

### Problem 33
To find the equation of the line of intersection of the two planes:

1. The first plane is $$ x + y + z - 6 = 0 $$.
2. The second plane is $$ 2x + 3y + 4z + 5 = 0 $$.

We can solve these equations simultaneously to find the line of intersection.

Let's express $$ z $$ in terms of $$ x $$ and $$ y $$ from the first equation:

$$
z = 6 - x - y
$$

Substituting this into the second plane's equation:

$$
2x + 3y + 4(6 - x - y) + 5 = 0
$$

This simplifies to:

$$
2x + 3y + 24 - 4x - 4y + 5 = 0 \implies -2x - y + 29 = 0 \implies y = -2x + 29
$$

Now substituting $$ y $$ back into the equation for $$ z $$:

$$
z = 6 - x - (-2x + 29) = 6 - x + 2x - 29 = x - 23
$$

Thus, the parametric equations for the line of intersection can be expressed as:

$$
x = t, \quad y = -2t + 29, \quad z = t - 23
$$

This can be rewritten in the form:

$$
x = t, \quad y = -2t + 29, \quad z = t - 23
$$

The final answer for the line of intersection is:

$$
x = t, \quad y = -2t + 29, \quad z = t - 23
$$

### Summary of Answers
- **30**: The point that does not lie on the same side of the plane is **(1, 2, 1)**.
- **31**: The value of $$ c $$ is **-3**.
- **32**: The angle between the line and the plane is **60 degrees**.
- **33**: The equation of the line of intersection is $$ x = t, y = -2t + 29, z = t - 23 $$.
- **30**: The point that does not lie on the same side of the plane is **(1, 2, 1)**. - **31**: The value of $$ c $$ is **-3**. - **32**: The angle between the line and the plane is **60 degrees**. - **33**: The equation of the line of intersection is $$ x = t, y = -2t + 29, z = t - 23 $$.

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