Question

The perpendicular distance from origin to the plane 3x + 4y + 5z = 25 is​

Answer 1

Step-by-step explanation:

Text Solution

350−−√350

650−−√650

62–√62

452–√452

Answer :

C

Answer 2

The perpendicular distance from origin to the plane 3x + 4y + 5z = 25 is 5/√2 unit

Given :

The equation of the plane 3x + 4y + 5z = 25

To find :

The perpendicular distance from origin to the plane

Solution :

Step 1 of 2 :

Write down the given equation of the plane

Step 2 of 2 :

Find perpendicular distance from origin to the plane

The given equation of the plane is

3x + 4y + 5z - 25 = 0

Origin represents the point (0,0,0)

Hence perpendicular distance from origin to the plane

$\displaystyle \sf{ = \bigg| \frac{(3 \times 0) + (4 \times 0) + (5 \times 0) - 25}{ \sqrt{ {3}^{2} + {4}^{2} + {5}^{2} } } \bigg| \: \: \: unit}$

$\displaystyle \sf{ = \bigg| \frac{0 + 0 + 0 - 25}{ \sqrt{9 + 16 + 25} } \bigg| \: \: \: unit}$

$\displaystyle \sf{ = \bigg| \frac{ - 25}{ \sqrt{50} } \bigg| \: \: \: unit}$

$\displaystyle \sf{ = \frac{25}{ \sqrt{50} } \: \: \: unit}$

$\displaystyle \sf{ = \frac{25}{ 5\sqrt{2} } \: \: \: unit}$

$\displaystyle \sf{ = \frac{5}{ \sqrt{2} } \: \: \: unit}$