Exercise 4(b)
Find the vector equation of the line passing through the point 2i +3j+k and parallel to the vector
4i-2j +

1 thought on “Exercise 4(b)<br />Find the vector equation of the line passing through the point 2i +3j+k and parallel to the vector<br />4i-2j +”

1. Answer:

   Given :-

   Ratio of length breadth and height = 1:2:3

   TSA = 792 sq. m

   To Find :-

   Volume

   Solution :-

   Let,

   Length = x

   breadth = 2x

   Height = 3x

   \bf\red{TSA = 2(lb + bh +lh)}TSA=2(lb+bh+lh)

   \sf 792 = 2(x\times 2x +2 x\times 3x+ x\times 3x)792=2(x×2x+2x×3x+x×3x)

   \sf 792=2(2x^{2} +6x^{2} +3x^{2} )792=2(2x

   2

   +6x

   2

   +3x

   2

   )

   \sf 792 = 2(11x^{2} )792=2(11x

   2

   )

   \sf\dfrac{792}{2} = 11x^2

   2

   792

   =11x

   2

   \sf 396 = 11x^{2}396=11x

   2

   \sf \dfrac{396}{11} = x^{2}

   11

   396

   =x

   2

   \sf 36 = x^{2}36=x

   2

   \sf \sqrt{36} =\sqrt{x^2}

   36

   =

   x

   2

   \sf 6 = x6=x

   Length = 6 m

   Breadth = 2(6) = 12 m

   Height = 3(6) = 18 m

   ~Finding volume

   \bf Volume = l\times b \times hVolume=l×b×h

   \sf Volume = 6 \times12\times18Volume=6×12×18

   \sf Volume = 1296 m^{3}Volume=1296m

   3

   Reply