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Vector Algebra

Short 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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Long 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 $fraction numerator negative 2 over denominator square root of 17 end fraction comma space fraction numerator negative 2 over denominator square root of 17 end fraction comma space fraction numerator 3 over denominator square root of 17 end fraction.$

Let O be the origin.
A is (2, -1, 3)

rightwards double arrow

OA 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 3 space straight k with hat on top

B is (8, 5, -6)

rightwards double arrow

OB with rightwards arrow on top space equals space 8 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space space 6 space straight k with hat on top

C is (4, 1, 0)

rightwards double arrow

OC with rightwards arrow on top space equals space 4 space straight i with hat on top space plus space straight j with hat on top space plus space 0 space straight k with hat on top space equals space 4 space straight i with hat on top space plus space straight j with hat on top

(i)

AB with rightwards arrow on top space equals space OB with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space open parentheses 8 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top close parentheses space minus space left parenthesis 2 space straight i with hat on top space minus space straight j with hat on top space plus space space 3 space straight k with hat on top right parenthesis space equals space 6 space straight i with hat on top space plus space 6 space straight j with hat on top space minus 9 space straight k with hat on top

therefore space space space space space AB space equals space square root of left parenthesis 6 right parenthesis squared plus left parenthesis 6 right parenthesis squared plus left parenthesis negative 9 right parenthesis squared end root space equals space square root of 36 plus 36 plus 81 end root space space space space space space space space space space space space space space space equals space square root of 153 space equals square root of 9 space cross times space 17 end root space equals space 3 space square root of 17

AC with rightwards arrow on top space equals space OC with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space left parenthesis 4 space straight i with hat on top space plus space straight j with hat on top right parenthesis space minus space left parenthesis 2 space straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 2 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top

therefore space space space AC space equals space square root of left parenthesis 2 right parenthesis squared plus left parenthesis 2 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared end root space equals space square root of 4 plus 4 plus 9 end root space equals space square root of 17 therefore space space AB space equals space 3 space space AC

(ii)

BC with rightwards arrow on top space equals space OC with rightwards arrow on top space minus space OB with rightwards arrow on top space equals space left parenthesis 4 space straight i with hat on top space plus space straight j with hat on top right parenthesis space minus space left parenthesis 8 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top right parenthesis space equals space minus 4 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 6 space straight k with hat on top

therefore space space space BC space equals space square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root space equals square root of 16 plus 16 plus 36 end root space equals square root of 68 equals space square root of 4 space cross times 17 end root space equals space 2 space square root of 17 therefore space space space space space straight a space unit space vector space along space BC with rightwards arrow on top space equals space fraction numerator BC with rightwards arrow on top over denominator open vertical bar BC with rightwards arrow on top close vertical bar end fraction space equals fraction numerator BC with rightwards arrow on top over denominator BC end fraction

equals space fraction numerator 1 over denominator 2 square root of 17 end fraction left parenthesis negative 4 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis space equals space minus fraction numerator 2 over denominator square root of 17 end fraction straight i with hat on top space minus space fraction numerator 2 over denominator square root of 17 end fraction straight j with hat on top space plus space fraction numerator 3 over denominator square root of 17 end fraction straight k with hat on top

therefore space space space

direction cosines of

BC with rightwards arrow on top

i.e., coeffs. of

straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top

are

negative fraction numerator 2 over denominator square root of 17 end fraction comma space minus fraction numerator 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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Short Answer Type

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.

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Long 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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Short 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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Long 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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