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Tentukan jarak antara
a. Titik pusat
O(0,0,0) dan garis
(x-4)/(3)=(y-2)/(4)=(4-z)/(5).
b. Garis
(x-2)/(-1)=(y-3)/(4)=(z)/(2) dan garis
(x+1)/(1)=(y-2)/(0)=(z+1)/(2)

$4$ . Tentukan jarak antara $\newline$ a. Titik pusat $O(0,0,0)$ dan garis $\frac{x-4}{3}=\frac{y-2}{4}=\frac{4-z}{5}$ . $\newline$ b. Garis $\frac{x-2}{-1}=\frac{y-3}{4}=\frac{z}{2}$ dan garis $\frac{x+1}{1}=\frac{y-2}{0}=\frac{z+1}{2}$

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Sum of finite series starts from 1

Full solution

Q.

4

. Tentukan jarak antara

\newline

a. Titik pusat

O(0,0,0)

dan garis

\frac{x-4}{3}=\frac{y-2}{4}=\frac{4-z}{5}

.

\newline

b. Garis

\frac{x-2}{-1}=\frac{y-3}{4}=\frac{z}{2}

dan garis

\frac{x+1}{1}=\frac{y-2}{0}=\frac{z+1}{2}

Find Direction Vector: For part a, we need to find the distance from the origin to the line. We use the formula for the distance from a point to a line in $3$ D: $d = \frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}}$ , where $Ax + By + Cz + D = 0$ is the equation of the plane.
Find Point on Line: First, we need to find the direction vector of the line, which is given by the coefficients of $(x-4)/3$ , $(y-2)/4$ , and $(4-z)/5$ . So the direction vector is $(3, 4, -5)$ .
Calculate Vector to Point: Next, we find a point on the line by setting the parameter to zero. If we set the parameter to zero, we get the point $(4, 2, 4)$ on the line.
Calculate Cross Product: Now we can find the vector from the origin to this point on the line, which is $(4-0, 2-0, 4-0) = (4, 2, 4)$ .
Simplify Cross Product: We calculate the cross product of the direction vector and the vector from the origin to the point on the line. The cross product is $\lvert(3, 4, -5) \times (4, 2, 4)\rvert$ .
Calculate Magnitude: The cross product is $(4*(-5) - 2*4, -5*4 - 3*4, 3*2 - 4*4) = (-20 - 8, -20 - 12, 6 - 16)$ .
Calculate Numerator: Simplifying the cross product, we get $(-28, -32, -10)$ .
Calculate Denominator: The magnitude of the cross product is $\sqrt{(-28)^2 + (-32)^2 + (-10)^2} = \sqrt{784 + 1024 + 100}$ .
Calculate Distance: Simplifying the magnitude, we get $\sqrt{1908}$ . This is the numerator of the distance formula.
Simplify Distance: The denominator of the distance formula is the magnitude of the direction vector, which is $\sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25}$ .
Simplify Distance: The denominator of the distance formula is the magnitude of the direction vector, which is $\sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25}$ . Simplifying the denominator, we get $\sqrt{50}$ .
Simplify Distance: The denominator of the distance formula is the magnitude of the direction vector, which is $\sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25}$ . Simplifying the denominator, we get $\sqrt{50}$ . The distance from the origin to the line is then $|\sqrt{1908}| / \sqrt{50}$ . This simplifies to $\sqrt{1908/50}$ .
Simplify Distance: The denominator of the distance formula is the magnitude of the direction vector, which is $\sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25}$ . Simplifying the denominator, we get $\sqrt{50}$ . The distance from the origin to the line is then $|\sqrt{1908}| / \sqrt{50}$ . This simplifies to $\sqrt{1908/50}$ . Simplifying the distance, we get $\sqrt{38.16}$ . This is incorrect because the square root of $1908$ divided by $50$ is not $38$ . $16$ . The correct calculation should be $\sqrt{1908/50} = \sqrt{38.16}$ .

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