Archive
8th of March 2017
Mathematics
Q:

Find the equation of the plane containing the parallel line x-1/2 = y+1/-1 = z/3 and x/4 =y-2/-2 = z+1/6. Also find if the plane thus obtained contains the line x-2/3 =y-1/1 = z-2/5


Yesterday i had asked question on maxima and minima .... A given quantity of metal sheet is to be cast ...... I wanted to get the answer with the help of a diagram

but you have only shown the diagram and not given the solution to the question . Kindly provide the complete solution to the question . This question is from RD SHARMA but

the same has not been solved in the solutions of RD SHARMA given on the site of topper learning. Please dont ask me to pose the question on Ask The Expert as this feature

is not working on the site of topper learning and is continuously showing General SQL Error. I have brought this to the notice of Customer Service of Topper Learning several times

but the issue has not yet been resolved . Till the issue is resolved , please provide complete solutions to all my questions

Ahuja8087 Ahuja - CBSE - Class XII Science

Wednesday, March 08, 2017 at 17:45:PM

A:

Question 1:

begin mathsize 12px style Recall comma space the space equation space of space straight a space plane space containing space two space parallel space lines.
fraction numerator straight x minus straight x subscript 1 over denominator straight l subscript 1 end fraction equals fraction numerator straight y minus straight y subscript 1 over denominator straight m subscript 1 end fraction equals fraction numerator straight z minus straight z subscript 1 over denominator straight n subscript 1 end fraction space and space fraction numerator straight x minus straight x subscript 2 over denominator straight l subscript 2 end fraction equals fraction numerator straight y minus straight y subscript 2 over denominator straight m subscript 2 end fraction equals fraction numerator straight z minus straight z subscript 2 over denominator straight n subscript 2 end fraction space is space given space by space open vertical bar table row cell straight x minus straight x subscript 1 end cell cell straight y minus straight y subscript 1 end cell cell straight z minus straight z subscript 1 end cell row cell straight l subscript 1 end cell cell straight m subscript 1 end cell cell straight n subscript 1 end cell row cell straight l subscript 2 end cell cell straight m subscript 2 end cell cell straight n subscript 2 end cell end table close vertical bar equals 0
Here comma space straight x subscript 1 equals 1 comma space straight y subscript 1 equals negative 1 comma space straight z subscript 1 equals 0 comma space straight x subscript 2 equals 0 comma space straight y subscript 2 equals 2 comma space straight z subscript 2 equals negative 1
straight l subscript 1 equals 2 comma space straight m subscript 1 equals negative 1 comma space straight n subscript 1 equals 3 comma space straight l subscript 2 equals 4 comma space straight m subscript 2 equals negative 2 comma space straight n subscript 2 equals 6
Substituting space in space open vertical bar table row cell straight x minus straight x subscript 1 end cell cell straight y minus straight y subscript 1 end cell cell straight z minus straight z subscript 1 end cell row cell straight l subscript 1 end cell cell straight m subscript 1 end cell cell straight n subscript 1 end cell row cell straight l subscript 2 end cell cell straight m subscript 2 end cell cell straight n subscript 2 end cell end table close vertical bar equals 0 space and space finding space the space determinant comma space you space will space the space required space equation. end style

Dear Student, I guess these days a little situation. We are extremely sorry. The second part to the above question will be added to you shortly before tomorrow's live chat. Thank you for cooperating. We will try to do avoid such mishaps henceforth.

With regards to your yesterday's question, the figure along with solution was given yesterday. There may be some technical issue because of which you were not able to view it. Hence, posting it again.

Wednesday, March 08, 2017 at 21:04:PM