AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

It is given that
X-intercept (a)=2
Y‌ -intercept ‌(b)=3
Z‌ -intercept ‌(c)=4
Therefore, the equation of the plane is,

x

2

+

y

3

+

z

4

=1
6x+4y+3z=12
The given points are A(−1,6,2),B(1,2,3) and C(−2,3,4)
Thus,
‌ DR's of ‌BC‌‌=(−2−1,3−2,4−3)
‌‌=(−3,1,1)
Therefore, the equation of plane passing through A(−1,6,2) and
DR′s(−3,1,1) is,
a(x−x1)+b(y−y1)+c(z−z1)‌‌=0
−3(x+1)+1(y−6)+1(z−2)‌‌=0
−3x+y+z−11‌‌=0
−3x+y+z‌‌=−11
Hence, the angle between the planes is
cos‌θ‌‌=

|(6)(−3)+(4)(1)+(3)(1)|

√36+16+9√9+1+1

‌‌=

|−18+4+3|

√61√11

‌‌=

√112

√61√11

‌‌=√

11

61

Solve further
θ=cos−1(√

11

61

)