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.

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.

82 Views

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.

76 Views

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 $PQ with rightwards arrow on top.$

80 Views

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.

74 Views

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.

77 Views

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).

77 Views

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.

Let $straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top$ be unit vectors along OA, OC, OE respectively. Let OB, OD, OF be the diagonals of faces OABC, OCDE, OEFA respectively of the cube of side of length b.

Now $OB with rightwards arrow on top space equals OA with rightwards arrow on top space plus space AB with rightwards arrow on top space equals space OA with rightwards arrow on top space plus space OC with rightwards arrow on top space equals space straight b space straight i with hat on top space plus space straight b space straight j with hat on top$
$therefore space space space open vertical bar OB with rightwards arrow on top close vertical bar space equals space square root of straight b squared plus straight b squared end root space equals space square root of 2 space straight b squared end root space equals space square root of 2 space straight b$
$therefore$ unit vector along
   $OB with rightwards arrow on top space equals space fraction numerator OB with rightwards arrow on top over denominator open vertical bar OB with rightwards arrow on top close vertical bar end fraction equals space fraction numerator straight b space straight i with hat on top space plus space straight b space straight j with hat on top over denominator square root of 2 space straight b end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis$
Similarly unit vector along $OD with rightwards arrow on top$ = $fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight j with hat on top plus straight k with hat on top right parenthesis$
and unit vector along $OF with rightwards arrow on top space equals space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight k with hat on top space plus space straight i with hat on top right parenthesis$
$therefore space space space space space OB with rightwards arrow on top space equals space fraction numerator straight a with rightwards arrow on top over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis comma space space space space OD with rightwards arrow on top space equals space fraction numerator 2 straight a over denominator square root of 2 end fraction left parenthesis straight j with hat on top space plus straight k with hat on top right parenthesis comma space space OF with rightwards arrow on top space equals space fraction numerator 3 straight a over denominator square root of 2 end fraction left parenthesis straight k with hat on top space plus space straight i with hat on top right parenthesis$
Let $straight R with rightwards arrow on top$ be resultant of vectors $OB with rightwards arrow on top comma space OD with rightwards arrow on top comma space OF with rightwards arrow on top$
$therefore space space space space space space space straight R with rightwards arrow on top space equals space OB with rightwards arrow on top space plus space OD with rightwards arrow on top space plus space OF with rightwards arrow on top space space space space space space space space space space space space space space space space equals fraction numerator straight a over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis space fraction numerator 2 straight a over denominator square root of 2 end fraction left parenthesis straight j with hat on top space plus space straight k with hat on top right parenthesis space plus space fraction numerator 3 straight a over denominator square root of 2 end fraction left parenthesis straight k with hat on top space plus space straight i with hat on top right parenthesis space space space space space space space space space space space space space space space space equals space fraction numerator 4 straight a over denominator square root of 2 end fraction straight i with hat on top space plus space fraction numerator 3 straight a over denominator square root of 2 end fraction straight j with hat on top space plus space fraction numerator 5 straight a over denominator square root of 2 end fraction straight k with hat on top$
$therefore$   magnitude of resultant vector = $open vertical bar straight R with rightwards arrow on top close vertical bar$
   $equals space square root of open parentheses fraction numerator 4 straight a over denominator square root of 2 end fraction close parentheses squared plus open parentheses fraction numerator 3 space straight a over denominator square root of 2 end fraction close parentheses squared plus space open parentheses fraction numerator 5 straight a over denominator square root of 2 end fraction close parentheses squared end root equals space square root of fraction numerator 16 straight a squared over denominator 2 end fraction plus fraction numerator 9 straight a squared over denominator 2 end fraction plus fraction numerator 25 space straight a squared over denominator 2 end fraction end root space equals space square root of 25 space straight a squared end root space equals space 5 space straight a$

263 Views

Short Answer Type

98. Write all the unit vectors in XY-plane.

149 Views

99. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

91 Views

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.

126 Views