The sum of distances from origin to (0, 5, 5) and (5, 8, 6) is:
5(√2 + √5)
To find the sum of distances from the origin to the given points, we need to calculate the individual distances first and then add them up. The origin in a 3D coordinate system is the point \((0, 0, 0)\).
The distance between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) in three-dimensional space is given by the distance formula:
\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Since we are calculating the distance from the origin \((0, 0, 0)\) to a point \((x, y, z)\), the formula simplifies to:
\[ D = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{x^2 + y^2 + z^2} \]
Let's find the distance from the origin \((0, 0, 0)\) to the first point \((0, 5, 5)\).
Here, \(x = 0, y = 5, z = 5\).
Using the simplified distance formula:
\[ D_1 = \sqrt{0^2 + 5^2 + 5^2} \]
\[ D_1 = \sqrt{0 + 25 + 25} \]
\[ D_1 = \sqrt{50} \]
To simplify \(\sqrt{50}\), we can factor out the largest perfect square, which is \(25\):
\[ D_1 = \sqrt{25 \times 2} \]
\[ D_1 = 5\sqrt{2} \]
Next, we find the distance from the origin \((0, 0, 0)\) to the second point \((5, 8, 6)\).
Here, \(x = 5, y = 8, z = 6\).
Using the simplified distance formula:
\[ D_2 = \sqrt{5^2 + 8^2 + 6^2} \]
\[ D_2 = \sqrt{25 + 64 + 36} \]
\[ D_2 = \sqrt{125} \]
To simplify \(\sqrt{125}\), we can factor out the largest perfect square, which is \(25\):
\[ D_2 = \sqrt{25 \times 5} \]
\[ D_2 = 5\sqrt{5} \]
The question asks for the sum of distances from the origin to both points. So, we add the two calculated distances, \(D_1\) and \(D_2\):
\[ \text{Sum} = D_1 + D_2 \]
\[ \text{Sum} = 5\sqrt{2} + 5\sqrt{5} \]
We can factor out the common term \(5\) from both terms:
\[ \text{Sum} = 5(\sqrt{2} + \sqrt{5}) \]
Therefore, the sum of distances from the origin to (0, 5, 5) and (5, 8, 6) is \(5(\sqrt{2} + \sqrt{5})\).
Determine the shortest distance between the lines l1 and l2 whose vector equations are
\(\vec{r}=2 \hat{\imath}+\hat{\jmath}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \quad \ldots(1)\)
\(\vec{r}=3 \hat{\imath}+\hat{\jmath}-\hat{k}+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \quad \ldots(2)\)
Find the distance between the points P (2, -1, 3) and Q (-5, 2, 1)?
The distance between the points (2, 3) and (4, 1) is.
The vertices of a triangle are \(A(2, 0, 0)\), \(B(0, 6, 0)\) and \(C(0, 0, 4)\). If \(AD\), \(BE\) and \(CF\) are the medians of the triangle, then what is \(AD^2 + BE^2 + CF^2\) equal to?
The distance of a point from the x-axis is called that point.