1answer for 1 given is only 9/17.
1. if DA=a, AB=b and CB=ka where k>0 and X,Y are the mid pts of DB and AC respt. such that |a|=17 and |XY|=4 then k is
a)8/17 b)9/17 c)25/17 d)4/17
2. ax(bxc), b(cxa), cx(axb) are
a)linearly dependent b)equal c)parallel d)none of these
3. a point O is the centre of a circle circumscribed about a triangle ABC. then OAsin2A + OBsin2B + OCsin 2C equals
a)(OA + OB + OC)sin2A
b)3OG where G is centroid of triangle
c)O
d)none of these
answers given are: 1.b 2.c 3.a
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4 Answers
211).......
\bg_white \100dpi Assume\: vertex\: C\: to\: be \:origin\:\vec{0},\:and \:vertices \:B\:, A\:, D\:, as\: \vec{B},\:\vec{A},\:\vec{D}\\So\:\vec{A}-\vec{D}=\vec{a}\\\vec{B}-\vec{A}=\vec{b} \\\vec{X}=\frac{\vec{D}+\vec{B}}{2}\\\\\vec{Y}=\frac{\vec{A}}{2}\\and\:we\:know\:\vec{B}-\vec{C}=k\vec{a}\\its\:given\left | \vec{a} \right |=17\:and\:\left | \vec{XY} \right | =4 \\we\: get\\\left | \frac{\vec{A}-\vec{D}-\vec{B}}{2} \right |=4\ \\\left | \vec{A}-\vec{D} \right |=17\\
\bg_white \100dpi Now\: sqauring\:we\:get\\ \left | {\vec{A}-\vec{D}} \right |^{2}+\left | \vec{B} \right |^{2}-2(\vec{A}-\vec{D}).\vec{B}=64\\ we\:know\:\vec{B}=k.\vec{a} \\and\ \vec{A}-\vec{D}=\vec{a}\\ so\: \left | \vec{a} \right|^2+k^2.\left | \vec{a} \right|^2-2k\vec{a}.\vec{a}=64\\ 17^2+k^2.17^2-2.k.17^2=64\\ 17^2(1+k^2-2k)=64\\ (k-1)^2=(\frac{8}{17})^2\\ \therefore k=9/17\: or\: 25/17
21nope... der will be two values of k!
cant u see it from soln abov...