Prove that the vectors forms the sides of a right-angled triang

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 QuestionsLong Answer Type

191. Show that the vectors 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top comma space straight b with rightwards arrow on top space equals space straight i with hat on top space minus space 3 straight j with hat on top space minus space 5 straight k with hat on top comma space straight c with rightwards arrow on top space equals space 3 straight i with hat on top minus 4 straight j with hat on top minus 4 straight k with hat on top form the vertices of a right angled triangle. 

80 Views

Advertisement

192. Prove that the vectors
straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top comma space space space space space straight b with rightwards arrow on top space equals space straight i with hat on top space minus 3 space straight j with hat on top space minus space 5 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 4 space straight j with hat on top space minus space 4 space straight k with hat on top forms the sides of a right-angled triangle. Also, find the remaining angles of the triangle.


Here
             straight a with rightwards arrow on top equals space 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top comma space straight b with rightwards arrow on top equals straight i with hat on top space minus 3 space straight j with hat on top space minus space 5 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 4 space straight j with hat on top space minus space 4 space straight k with hat on top
Now,    straight a with rightwards arrow on top plus straight b with rightwards arrow on top space equals space left parenthesis 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top right parenthesis plus left parenthesis straight i with hat on top minus 3 space straight j with hat on top minus 5 space straight k with hat on top right parenthesis equals 3 space straight i with hat on top minus 4 space straight j with hat on top minus 4 space straight k with hat on top
therefore space space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top space are space coplanar space vectors space and space form space the space sides space of space straight a space triangle.
Now,   straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space left parenthesis 2 right parenthesis space left parenthesis 1 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis negative 3 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis negative 5 right parenthesis space equals space 2 plus 3 minus 5 space equals space 0


therefore space space space space straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space are space perpendicular
therefore space space space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top space form space the space sides space of space straight a space right minus angled space triangle. space
space space space space space space space space space space space cos space straight A space equals space fraction numerator straight b with rightwards arrow on top. straight c with rightwards arrow on top over denominator open vertical bar straight b with rightwards arrow on top close vertical bar. space open vertical bar straight c with rightwards arrow on top close vertical bar end fraction
space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis 1 right parenthesis left parenthesis 3 right parenthesis plus left parenthesis negative 3 right parenthesis thin space left parenthesis negative 4 right parenthesis space plus space left parenthesis negative 5 right parenthesis thin space left parenthesis negative 4 right parenthesis over denominator square root of 1 plus 9 plus 25 end root space square root of 9 plus 16 plus 6 end root end fraction
space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 3 plus 12 plus 20 over denominator square root of 35 space square root of 41 end fraction space equals fraction numerator 35 over denominator square root of 35 space square root of 41 end fraction
space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator square root of 35 over denominator square root of 41 end fraction space equals square root of 35 over 41 end root
therefore space space space space space space space straight A space equals space cos to the power of negative 1 end exponent open parentheses square root of 35 over 41 end root close parentheses
space space space space space space space space space cos space straight B space equals space fraction numerator straight c with rightwards arrow on top. space straight a with rightwards arrow on top over denominator open vertical bar straight c with rightwards arrow on top close vertical bar space open vertical bar straight a with rightwards arrow on top close vertical bar end fraction space equals fraction numerator left parenthesis 3 right parenthesis left parenthesis 2 right parenthesis plus left parenthesis negative 4 right parenthesis thin space left parenthesis negative 1 right parenthesis plus left parenthesis negative 4 right parenthesis thin space left parenthesis 1 right parenthesis over denominator square root of 9 plus 16 plus 16 end root space square root of 4 plus 1 plus 1 end root end fraction
space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 6 plus 4 minus 4 over denominator square root of 41 space square root of 6 end fraction space equals space fraction numerator square root of 6 over denominator square root of 41 space square root of 6 end fraction space equals space fraction numerator square root of 6 over denominator square root of 41 end fraction equals space square root of 6 over 41 end root
because space space space space straight B space equals space cos to the power of negative 1 end exponent space open parentheses square root of 6 over 41 end root close parentheses









73 Views

Advertisement

 Multiple Choice QuestionsShort Answer Type

193. Show that the points whose position vectors are 
straight a with rightwards arrow on top space equals space 4 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space space 5 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals straight i with hat on top space minus space straight j with hat on top form a right angled triangle.
75 Views

194.

If the vertices A, B, C of ∆ ABC have position vectors (1, 2. 3), (– 1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

95 Views

Advertisement
195.

Show that 1 over 7 left parenthesis 2 straight i with hat on top space plus space 3 straight j with hat on top space plus space 6 straight k with hat on top right parenthesis comma space space 1 over 7 left parenthesis 3 straight i with hat on top minus 6 straight j with hat on top plus 2 straight k with hat on top right parenthesis comma space space 1 over 7 left parenthesis 6 straight i with hat on top plus 2 straight j with hat on top minus 3 straight k with hat on top right parenthesis are mutually perpendicular unit vectors.

74 Views

 Multiple Choice QuestionsLong Answer Type

196. The scalar product of the vector straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top with a unit vector along the sum of vectors 2 straight i with hat on top space plus space 4 space straight j with hat on top space minus space 5 straight k with hat on top and straight lambda straight i with hat on top space plus space 2 straight j with hat on top space plus space space 3 straight k with hat on top is equal to 1. Find the value of  λ.
76 Views

197. If a unit vector straight a with rightwards arrow on top makes angle straight pi over 4 space with space straight i with hat on top comma space space straight pi over 3 space with space straight j with hat on top and an acute angle straight theta with straight k with hat on top, then find the component of straight a with rightwards arrow on top and the angle straight theta.
80 Views

 Multiple Choice QuestionsShort Answer Type

198. If A, B, C have position vectors (0, 1, 1) (3, 1, 5), (0. 3, 3) respectively, then show that the ∆ABC is right angled at C.
74 Views

Advertisement

 Multiple Choice QuestionsLong Answer Type

199. If a unit vector straight a with rightwards arrow on top makes angles straight pi over 3 with straight i with hat on top comma  straight pi over 4 space with space straight j with hat on top and acute angle straight theta with straight k with hat on top, then find the components of straight a with rightwards arrow on top and the angle straight theta.
73 Views

200.

Let straight a with rightwards arrow on top space equals space 4 straight i with hat on top space plus space 5 straight j with hat on top space minus space straight k with hat on top comma space straight b with rightwards arrow on top space equals space straight i with hat on top minus 4 straight j with hat on top plus space 5 straight k with hat on top space and space straight c with rightwards arrow on top space equals 3 straight i with hat on top plus straight j with hat on top minus straight k with hat on top. space Find a vector straight d with rightwards arrow on top which is perpendicular to both straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space and space straight d with rightwards arrow on top. space straight c with rightwards arrow on top space equals space 21.

75 Views

Advertisement