The scalar product of the vector  with a unit vector along th

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 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. 

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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.
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 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.
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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?

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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.

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 Multiple Choice QuestionsLong Answer Type

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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.


Let straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top
and straight b with rightwards arrow on top space equals space unit space vector space parallel space to space the space sum space of space vectors space 2 straight i with hat on top space plus space 4 straight j with hat on top space minus 5 straight k with hat on top comma space space straight lambda straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top
         equals space fraction numerator left parenthesis 2 plus straight lambda right parenthesis space straight i with hat on top space plus space 6 space straight j with hat on top space minus space 2 space straight k with hat on top over denominator square root of left parenthesis 2 plus straight lambda right parenthesis squared plus left parenthesis 6 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction space equals space fraction numerator 1 over denominator square root of straight lambda squared plus 4 straight lambda plus 44 end root end fraction space equals space open curly brackets left parenthesis 2 plus straight lambda right parenthesis straight i with hat on top space plus space space 6 space straight j with hat on top space minus space 2 space straight k with hat on top close curly brackets
Now,  straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 1
therefore space space space space space fraction numerator 1 over denominator square root of straight lambda squared plus 4 straight lambda plus 44 end root end fraction space left square bracket left parenthesis 1 right parenthesis plus space left parenthesis 2 plus straight lambda right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 6 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis negative 2 right parenthesis right square bracket space equals space 1
therefore space space space space space straight lambda plus 6 space equals square root of straight lambda squared plus 4 straight lambda plus 44 end root
rightwards double arrow space space space space straight lambda squared plus 12 straight lambda space plus 36 space equals space straight lambda squared plus 4 straight lambda plus 44
rightwards double arrow space space space space space space space 8 straight lambda space equals space 8 space space space space space space rightwards double arrow space space space space space straight lambda space equals space 1
straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top
2 straight i with hat on top space plus space 4 straight j with hat on top space minus space 5 space straight k with hat on top
space space space space space space space straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top
and space straight b with rightwards arrow on top space equals space unit space vector space parallel space to space the space sum space of space vectors space 2 straight i with hat on top space plus space 4 straight j with hat on top minus space 5 straight k with hat on top comma space straight lambda straight i with hat on top space plus space 2 straight j with hat on top space plus space 3 straight k with hat on top space space space
space space space space space space space space space space space space equals space fraction numerator left parenthesis 2 plus straight lambda right parenthesis space straight i space plus space 6 space straight j with hat on top space minus space 2 space straight k with hat on top over denominator square root of left parenthesis 2 plus straight lambda squared right parenthesis plus left parenthesis 6 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction
space space space space space space space space space space space equals space fraction numerator 1 over denominator square root of straight lambda squared plus 4 straight lambda plus 44 end root end fraction open curly brackets left parenthesis 2 plus straight lambda right parenthesis space straight i with hat on top space plus space 6 space straight j with hat on top space minus space 2 space straight k with hat on top close curly brackets space
Now space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis given right parenthesis
therefore space space fraction numerator 1 over denominator square root of straight lambda squared plus 4 straight lambda plus 44 end root end fraction left square bracket left parenthesis 1 right parenthesis space plus space left parenthesis 2 plus straight lambda right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 6 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis negative 2 right parenthesis right square bracket space equals space 1
rightwards double arrow space space space space straight lambda squared plus 12 straight lambda plus 36 space equals space straight lambda squared plus 4 straight lambda plus 44 space
rightwards double arrow space space space space 8 space straight lambda space equals space 8 space space space space space space space space space space space space space space space space space space space rightwards double arrow space space straight lambda space equals space 1

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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.
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 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.
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 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.
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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.

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