AskSia

Plus

18-20 題為題組坐標空間中, 設

O

為原點,

E

為平面

x-z=4

。試回答下列問題。 18. 若原點

O

在平面

E

上的投影...

Jul 28, 2024

Solution by Steps

step 1

Identify the plane equation

x - z = 4

and the origin

O(0,0,0)

step 2

Find the projection point

Q

O

on the plane

E

. The projection of a point on a plane can be found using the formula for the perpendicular distance

step 3

The normal vector to the plane

x - z = 4

\vec{n} = (1, 0, -1)

step 4

The projection point

Q

can be found by solving the equation of the plane with the normal vector. Since

O

is the origin, the projection

Q

will be along the direction of the normal vector

step 5

The coordinates of

Q

can be found by solving

x - z = 4

with

x = z

. This gives

Q(2, 0, -2)

step 6

Calculate the vector

\overrightarrow{OQ} = Q - O = (2, 0, -2)

step 7

Find the angle

\alpha

between

\overrightarrow{OQ}

and

(1,0,0)

using the dot product formula:

\cos \alpha = \frac{\overrightarrow{OQ} \cdot (1,0,0)}{|\overrightarrow{OQ}| \cdot |(1,0,0)|}

step 8

Compute the dot product:

\overrightarrow{OQ} \cdot (1,0,0) = 2

step 9

Compute the magnitudes:

|\overrightarrow{OQ}| = \sqrt{2^2 + 0^2 + (-2)^2} = 2\sqrt{2}

and

|(1,0,0)| = 1

step 10

Substitute into the formula:

\cos \alpha = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

step 11

The correct answer is (4)

\frac{\sqrt{2}}{2}

Answer

(4)

\frac{\sqrt{2}}{2}

Key Concept

Projection of a point on a plane

Explanation

The projection of a point on a plane can be found using the normal vector to the plane and the dot product formula.

Question 19

step 1

Given the point

P(a, b, c)

and the condition

\theta \leq \frac{\pi}{6}

, where

\theta

is the angle between

\overrightarrow{OP}

and

(1,0,0)

step 2

Use the dot product formula:

\cos \theta = \frac{\overrightarrow{OP} \cdot (1,0,0)}{|\overrightarrow{OP}| \cdot |(1,0,0)|}

step 3

Since

\theta \leq \frac{\pi}{6}

\cos \theta \geq \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}

step 4

Substitute into the dot product formula:

\frac{a}{\sqrt{a^2 + b^2 + c^2}} \geq \frac{\sqrt{3}}{2}

step 5

Square both sides:

\left(\frac{a}{\sqrt{a^2 + b^2 + c^2}}\right)^2 \geq \left(\frac{\sqrt{3}}{2}\right)^2

step 6

Simplify:

\frac{a^2}{a^2 + b^2 + c^2} \geq \frac{3}{4}

step 7

Cross-multiply:

4a^2 \geq 3(a^2 + b^2 + c^2)

step 8

Simplify:

4a^2 \geq 3a^2 + 3b^2 + 3c^2

step 9

Rearrange:

a^2 \geq 3(b^2 + c^2)

Answer

a^2 \geq 3(b^2 + c^2)

Key Concept

Dot product and angle between vectors

Explanation

The dot product formula can be used to find the relationship between the components of vectors and the angle between them.

Question 20

step 1

Given

P

is on the plane

E

and

b = 0

. The plane equation is

x - z = 4

step 2

Substitute

b = 0

into the inequality from Question 19:

a^2 \geq 3c^2

step 3

Since

P

is on the plane

E

a - c = 4

step 4

Substitute

a = c + 4

into the inequality:

(c + 4)^2 \geq 3c^2

step 5

Expand and simplify:

c^2 + 8c + 16 \geq 3c^2

step 6

Rearrange:

16 \geq 2c^2 - 8c

step 7

Solve the quadratic inequality:

2c^2 - 8c - 16 \leq 0

step 8

Factorize:

2(c^2 - 4c - 8) \leq 0

step 9

Solve for

c

c^2 - 4c - 8 = 0

. The roots are

c = 2 \pm 2\sqrt{3}

step 10

The range for

c

2 - 2\sqrt{3} \leq c \leq 2 + 2\sqrt{3}

step 11

The minimum length of

\overline{OP}

is when

c = 2 - 2\sqrt{3}

, giving

a = 4 - 2\sqrt{3}

. The length is

\sqrt{(4 - 2\sqrt{3})^2 + 0 + (2 - 2\sqrt{3})^2}

step 12

Simplify:

\sqrt{(4 - 2\sqrt{3})^2 + (2 - 2\sqrt{3})^2} = \sqrt{16 - 16\sqrt{3} + 12} = 2

Answer

2 - 2\sqrt{3} \leq c \leq 2 + 2\sqrt{3}

, minimum length is

2

Key Concept

Quadratic inequalities and distance in 3D space

Explanation

Solving quadratic inequalities helps find the range of possible values, and the distance formula in 3D space helps find the minimum length.