How can I find the equation of the plane passing through a point and perpendicular to a line?
As the line is perpendicular to the plane the direction ratios or the cosine ratios of the direction of plane will be same as that of the line but as the plane is passing through a particular point, putting that point in the equation and solving for the value of the constant term, the equation can be found out.For example;let the plane is perpendicular to the lineaX+bY+cZ+d=0then the equation of the plane will be some what likeaX+bY+cZ+D=0where the a,b,c are same as that of the line and the only unknown quantity is the “D.”As the line is passing through a point, lets say (p,q,r) then it must satisfy the eqn of the plane. Thus putting the point in the eqn we get;ap+bq+cr+D=0in this eqn only unknown quantity is D and the value of D can be found from the above eqn and putting the value of D in the parent eqn the eqn of Plane can be found out.
Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED have a length of 2cm , 6cm and 3 cm respectively. What is the length of diameter of a circle in cm?
We need to use two geometry properties to answer this:(1) When two chords intersect inside a circle, the product of the two lengths on one chord = the product of the two lengths on the other chord.So AE * EB = CE * ED2 * 6 = 4 * ED,12 = 4EDED = 3.So now we know the lengths of all 4 of the little segments on the chords.(2) The center of a circle is on the perpendicular bisector of any chord.Let’s locate the chords conveniently on an x/y axis. We can put E at the origin, and the chords will fall along the x and y axes. Locate A and B on the x-axis and C and D on the y-axis withA = (-2, 0) and B = (6, 0)C = (0, 4) and D = (0, -3)Now you can easily find the midpoints of the two chord by averaging the endpoints (or just looking at your graph).AB: (2, 0) CD : (0, .5)Since the chords are vertical and horizontal, their intersection is easy to see. The intersection must be the center of the circle.Center = (2, .5)The length r of the radius would be the distance from the Center to any of the four points A, B, C, or D. We can use the Pythagorean theorem.Let’s use point C: (0, 4)r^2 = (2–0)^2 + (4 - .5)^2r^2 = 4 + 49/4 = ( 16 + 49)/ 4 = 65 / 4So r =( Sqrt65)/ 2The diameter is twice that: d = Sqrt 65You can confirm this by doing the same calculation with A, B, or D.
An airplane has a ground speed of 350 km/hr in the direction due west. If there is a wind blowing northwest at 40 km/hr, calculate the true air speed and heading of the airplane?
Hello my friend. Thanks for the A2A…Airspeed is the speed of the aircraft relative to the wind blowing around it.Ground speed is the speed of the aircraft relative to the ground.There will be higher ground speed and low true airspeed if there is a tailwind. There will be low ground speed and higher airspeed is a headwind.Now, Ground speed = True Airspeed + Windspeed. So,True Airspeed = Ground speed — Wind speed.Aeronautical Calculations of Headwind and Crosswind.Here it is given that the aircraft is moving towards west. Therefore it is heading 270 degrees at a Groundspeed of 350 kmph.Now, the wind is blowing from Northwest.It means it is blowing from 315 degrees. The wind is blowing from NW to SE.So, the aircraft has both headwind component as well as cross wind components.So, angle between the wind and aircraft is therefore 315 — 270 = 45 degrees.It is to be remembered thatHeadwind = Windspeed * Cos (A).Crosswind = Windspeed * Sin (A).So, here, the headwind will beHW = 40 * Cos 45 = 40 * 0.707 = 28.28 km/h.CW = 40 * Sin 45 = 40* 0.707 =28.28 km/h.True Airspeed (TAS) =350 — ( — 28.28) = 350+28.28 = 378.28 km/h.Windspeed will be negative as it is a headwind. It will be positive as it is a tailwind.Due to the 45 degrees of cross wind, the track of the airplane will change to the below calculated Drift Angle.Drift Angle = (Windspeed * Angle between aircraft & wind) / True Airspeed.Drift angle = (40 *45)/378.28 = 4.75 degrees towards the Wind.So, the drift angle will be also 4.75 degrees.So, the new heading of the aircraft will be 274.75 degrees (West north-West); the true airspeed of the aircraft will be 378.28 km/h.See that the groundspeed is less than the true airspeed. So, it is headwind.Hope you got your answer. Thanks a lot..
ALGEBRA 2: How do I write an equation of a line when I only know the x-intercept & the y-intercept?
Here... 1) if you have a y intercept, it crosses the y axis, thus x is zero. The opposite applies for x intercepts--- in an x intercept, the y is zero. So the 2 points would be (0, -2.1) and (3.5, 0). You find the line of that.. u know how? Basically, take the m which equals the y of point two which happens to be 0 and the y of point one and divide it by the difference of the x of point two minus the x of point one. THe equation would be y= (the slope)x +(the y intercept) 2) You write out the two points. (1.2, 5.1) and (3.7, 0) and find the equation. Hope that helped!
A taxi firm charges a fixed amount plus so much per mile. A journey of 6 miles costs $3.70. A journey of 10*?
so money per mile=m 6m=3.7 10m=5.1 solve for 8m you add the two equations together... yes ADD, that is the easy way instead of solving for "m" so 16m=8.8 and divide by 2 8m=4.4 $4.40