1
AIEEE 2009
+4
-1
Out of Syllabus
Let the line $$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane $$\,\,\,\,\,$$ $$x + 3y - \alpha z + \beta = 0.$$ Then $$\left( {\alpha ,\beta } \right)$$ equals
A
$$(-6,7)$$
B
$$(5,-15)$$
C
$$(-5,5)$$
D
$$(6, -17)$$
2
AIEEE 2009
+4
-1
The projections of a vector on the three coordinate axis are $$6,-3,2$$ respectively. The direction cosines of the vector are :
A
$${6 \over 5},{{ - 3} \over 5},{2 \over 5}$$
B
$${6 \over 7 },{{ - 3} \over 7},{2 \over 7}$$
C
$${- 6 \over 7 },{{ - 3} \over 7},{2 \over 7}$$
D
$$6, -3, 2$$
3
AIEEE 2008
+4
-1
The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then
A
$$a=2,$$ $$b=8$$
B
$$a=4,$$ $$b=6$$
C
$$a=6,$$ $$b=4$$
D
$$a=8,$$ $$b=2$$
4
AIEEE 2008
+4
-1
If the straight lines $$\,\,\,\,\,$$ $$\,\,\,\,\,$$ $${{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ $$\,\,\,\,\,$$ and$$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}$$ intersects at a point, then the integer $$k$$ is equal to
A
$$-5$$
B
$$5$$
C
$$2$$
D
$$-2$$
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