Show that the angle between two diagonals of a cube is  from M

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsShort Answer Type

221.
Prove that an angle inscribed in a semi-circle is a right angle.
87 Views

 Multiple Choice QuestionsLong Answer Type

222. Using vectors, prove that if two medians of a triangle ABC be equal, then it is an isosceles triangle. 
157 Views

223. Using vectors, prove that the perpendicular bisectors of the sides of a triangle are concurrent.
274 Views

 Multiple Choice QuestionsShort Answer Type

224. Prove that the median to the base of isosceles triangle is perpendicular to the base.
257 Views

Advertisement

 Multiple Choice QuestionsLong Answer Type

225. If the median of the base of a triangle is perpendicular on the base, then prove that the triangle is an isosceles.
90 Views

226. (i) Prove that cos (α + β) )= cos α cos β – sin α sin β.
(ii)  Prove that cos (α – p) = cos α cos β + sin α sin β.
85 Views

227. Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
87 Views

 Multiple Choice QuestionsShort Answer Type

228. Using vectors, prove that a parallelogram whose diagonals are equal is a rectangle. 
88 Views

Advertisement

 Multiple Choice QuestionsLong Answer Type

229. Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.
78 Views

Advertisement

230. Show that the angle between two diagonals of a cube is cos to the power of negative 1 end exponent space open parentheses 1 third close parentheses.


Take O, a comer of cube OBLCMANP, as origin and OA, OB, OC. the three edges through it as the axes.

Let OA = OB = OC = α, then the co-ordinates of O, A, B, C are (0, 0, 0),
(a, 0. 0), (0, a, 0). (0, 0, a) respectively ; those of P, L, M, N are (a, a, a),(0, a. a), (a, 0, a), (a, a, 0) respectively.
The four diagonals are
OP, AL, BM, CN.
Consider the diagonals AL and BM
AL with rightwards arrow on top space equals space straight P. straight V. space of space straight L space minus space straight P. straight V. space of space straight A space equals space left parenthesis straight a space straight j with hat on top space plus space straight a space straight k with hat on top right parenthesis space minus space left parenthesis straight a space straight i with hat on top right parenthesis space equals space minus straight a space straight i with hat on top space plus straight a space straight j with hat on top space plus space straight a space straight k with hat on top
BM with rightwards arrow on top space equals space straight P. straight V. space of space straight M space minus space straight P. straight V. space of space straight B space equals space left parenthesis straight a straight i with hat on top space plus space straight a space straight k with hat on top right parenthesis space minus space left parenthesis straight a space straight j with hat on top right parenthesis space equals space straight a space straight i with hat on top space minus space straight a space straight j with hat on top space plus space straight a space straight k with hat on top
Let  straight theta be angle between AL and BM.
therefore space space space cos space straight theta space equals space fraction numerator AL with rightwards arrow on top space BM with rightwards arrow on top over denominator open vertical bar AL with rightwards arrow on top close vertical bar space open vertical bar BM with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator left parenthesis negative straight a right parenthesis thin space left parenthesis straight a right parenthesis space plus space left parenthesis straight a right parenthesis thin space left parenthesis negative straight a right parenthesis space plus space left parenthesis straight a right parenthesis space left parenthesis straight a right parenthesis over denominator square root of straight a squared plus straight a squared plus straight a squared end root space square root of straight a squared plus straight a squared plus straight a squared end root end fraction
space space space space space space space space space space space space space space space space space equals space fraction numerator negative straight a squared minus straight a squared plus straight a squared over denominator square root of 3 space straight a squared end root space square root of 3 straight a squared end root end fraction space equals space minus fraction numerator straight a squared over denominator 3 straight a squared end fraction space equals space minus 1 third
therefore space space space acute space angle space straight theta space is space given space by
                              cos space straight theta space equals space 1 third
therefore space space space straight theta space equals space cos to the power of negative 1 end exponent space open parentheses 1 third close parentheses
Similarly the angle between the other pairs of diagonals is cos to the power of negative 1 end exponent open parentheses 1 third close parentheses.
82 Views

Advertisement
Advertisement