Show that vector  is equally inclined to the axes OX. OY and O

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 Multiple Choice QuestionsShort Answer Type

91. Write the direction ratio’s of the vector 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 minus space 2 space straight k with hat on top and hence calculate its direction cosines.
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92. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction.
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93.

If straight P with rightwards arrow on top left parenthesis 1 comma space 5 comma space 4 right parenthesis and straight Q with rightwards arrow on top space left parenthesis 4 comma space 1 comma space minus 2 right parenthesis, find the direction ratios of PQ with rightwards arrow on top.

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 Multiple Choice QuestionsLong Answer Type

94. If A (2, – 1, 3), B (8, 5, – 6) and C (4, 1, 0) are the vertices of a triangle. show that
(i) AB = 3 AC 
(ii) The direction-cosines of BC with rightwards arrow on top are fraction numerator negative 2 over denominator square root of 17 end fraction comma space fraction numerator negative 2 over denominator square root of 17 end fraction comma space fraction numerator 3 over denominator square root of 17 end fraction.

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 Multiple Choice QuestionsShort Answer Type

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95. Show that 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 is equally inclined to the axes OX. OY and OZ. 


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 space straight k with hat on top

 therefore space space space space space straight a with rightwards arrow on top space equals space square root of left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared end root space equals space square root of 1 plus 1 plus 1 end root space equals square root of 3
therefore space space space space straight a with hat on top space equals space fraction numerator straight a over denominator open vertical bar straight a with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 3 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top right parenthesis space equals space fraction numerator 1 over denominator square root of 3 end fraction straight i with hat on top space plus space fraction numerator 1 over denominator square root of 3 end fraction straight j with hat on top space plus space fraction numerator 1 over denominator square root of 3 end fraction straight k with hat on top
therefore space space space space direction space cosines space of space straight a with rightwards arrow on top space are space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction
therefore space space space space cos space straight alpha space equals fraction numerator 1 over denominator square root of 3 end fraction comma space space cos space straight beta space equals space fraction numerator 1 over denominator square root of 3 end fraction comma space space cos space straight gamma space equals space fraction numerator 1 over denominator square root of 3 end fraction
where straight alpha comma space straight beta comma space straight gamma are angles made by vector with axes.
therefore space space space space cos space straight alpha space equals space cos space straight beta space equals space cos space straight gamma
rightwards double arrow space space space space space space space space space straight alpha space equals straight beta space equals straight gamma
therefore space space space space given space vector space is space equally space inclined space to space the space axes space OX comma space OY comma space OZ.

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 Multiple Choice QuestionsLong Answer Type

96. Show that the points A (6. – 7, 0) B (16, – 19, – 4). C (0, 3, – 6), D (2, – 5, 10) are such that AB and CD intersect at the point P (1, – 1, 2).
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97. The vectors of magnitude a, 2a, 3a meet at a point and their directions arc. along the diagonals of three adjacent faces of a cube. Determine the magnitude of their resultant.
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 Multiple Choice QuestionsShort Answer Type

98. Write all the unit vectors in XY-plane.
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99. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
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 Multiple Choice QuestionsLong Answer Type

100. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
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