Explanation
If a ray makes angles α,β,γ and δ with the four diagonals of a cube.
Then we know that cos2α+cos2β+cos2γ+cos2δ=43
A:
cos2α+cos2β+cos2γ+cos2δ=43
B:
sin2α+sin2β+sin2γ+sin2δ
=1−cos2α+1−cos2β+1−cos2γ+1−cos2δ=4−43=83
C:
cos2α+cos2β+cos2γ+cos2δ
=1−2sin2α+1−2sin2β+1−2sin2γ+1−2sin2δ=4−163=−43
Descending order is B,A,C
Hence, option A is correct answer.
Given lines are →r=3i+2j−4k+λ(i+2j+2k)
and →r=5i−2k+μ(3i+2j+6k)
We know, angle between →r=→a1+λ→b1 and →r=→a2+λ→b2 is given by,
cosθ=→b1.→b2|→b1||→b2|
⇒cosθ=(i+2j+2k).(3i+2j+6k)√12+22+22√32+22+62
=3+4+12√9√49=1921
⇒θ=cos−1(1921)
Let l,m,n be the direction ratio of a line making angles ,,, with four diagonals of a cube then as four diagonals have direction ratio,(a,a,a);(a,a,−a);(a,−a,a,) ;(−a,a,a)(where a is the side of the cube)
Now, cosα=|al+am+an√a2+a2+a2√l2+m2+n2|=|l+m+n√3√l2+m2+n2|
cosβ=|al+am−an√a2+a2+a2√l2+m2+n2|=|l+m−n√3√l2+m2+n2|
cosγ=|al−am+an√a2+a2+a2√l2+m2+n2|=|l−m+n√3√l2+m2+n2|
cosδ=|al+am+an√a2+a2+a2√l2+m2+n2|=|−l+m+n√3√l2+m2+n2|
and cos2α+cos2β+cos2γ+cos2δ
=13×(4l2+4m2+4n2)l2+m2+n2=43
So statement 1 is true and statement-2 is not true .
Hence, option 'C' is correct.
If the straight lines x−12=y+1K=z2 and x+15=y+12=zK are coplanar, then the plane(s) containing these two lines is(are)
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