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Q1.  If the coordinate of the verities of a triangle ABC be A-1,3,2, B2,3,5and C3,5,-2, then ∠ A is equal to
•  45°
•   60°
•   30°
•  90°
Solution
Now,AB=√(3^2+0+3^2 )=√18 CA=√(16+4+16)=6 and BC=√(1+4+49)=√54 ∵ 〖AB〗^2+〖CA〗^2=〖BC〗^2 ∴ ∆ABC is right angled triangle, right angled at, A.. Thus, ∠ A=90° .

Q2. The equation of the plane passing through the intersection of the planes x+2y+3z+4=0 and 4x+3y+2z+1=0 and the origin, is :
•  3x+2y+z+1=0
•  3x+2y+z=0
•  2x+3y+z=0
•   x+y+z=0
Solution
Equation of plane passing through the intersection of given planes, is (x+2y+3z+4)+Î»(4x+3y+2z+1)=0 …(i) Plane (i) is passing through the origin ie,(0,0,0) ∴4+Î»=0 ⇒ Î»=-4 On putting the value of Î» in Eq. (i), we get (x+2y+3z+4)-4(4x+3y+2z+1)=0 ⇒ -15x-10y-5z=0 ⇒3x+2y+z=0 .

Q3.   Which of the following is an equation of a sphere? :
•    x2+y2+z2-2xy-2yz-6zx=4
•  x2+y2+z2+1=0
•  x2+y2+z2-4x-4y-4z+25=0
•  x2+y2+z2-2x-2y-2z+2=0
Solution
Clearly in option (a), it is not a sphere as it contains xy,yz and zx terms. In options (b) and (d) u^2+v^2+w^2-c^2<0 .="" c="" is="" option="" so="" span="" sphere="">

Q4.  The value of k so that the lines x-1-3=y-22k=z-32 and, x-13k=y-11=z-6-5 may be perpendicular is given by
•   -7/10
•  10/7
•   -10
•  -10/7
Solution
Given lines will be perpendicular, if -3×3k+2k×1+2×-5=0⇒-7k-10=0⇒k=-10/7 .

Q5. If l1,m1,n1 and l2,m2,n2 are direction cosines of the two lines inclined to each other at an angle , then the direction cosines of the external bisector of the angle between the lines are
•   l1-l22sin Î¸/2 , m1-m22sin Î¸/2 , n1-n22sin Î¸/2
•  l1+l22cos Î¸/2 , m1+m22cos Î¸/2 , n1+n22cos Î¸/2
•  l1+l22sin Î¸/2 , m1+m22sin Î¸/2 , n1+n22sin Î¸/2
•  l1-l22cos Î¸/2 , m1-m22cos Î¸/2 , n1-n22cos Î¸/2
Solution
In Fig. OE is the external bisector The co-ordinates of E are ((l_1-l_2)/2,(m_1-m_2)/2,(n_1-n_2)/2) Therefore, direction ratios of OE are proportional to (l_1-l_2)/2,(m_1-m_2)/2,(n_1-n_2)/2 .

Q6.  Equation of the plane passing through the point (1, 1, 1) and perpendicular to each of the planes x+2y+3z=7 and 2x-3y+4z=0, is
•   17x-2y+7z=12
•  17x+2y-7z=12
• 17x+2y+7z=12
•   17x-2y-7z=12
Solution
The equation of the plane containing the line (x+1)/(-3)=(y-3)/2=(z+2)/1 is a(x+1)+b(y-3)+c(z+2)=0 …(i) Where, -3a+2b+c=0 …(ii) This passes through (0,7,-7) ∴a+4b-5c=0 From (ii) and (iii), we have a/(-14)=b/(-14)=c/(-14)⇒a/1=b/1=c/1 So, the required plane is x+y+z=0 .

Q7. The equation of a line is 6x-2=3y-1=2z-2 The direction ratios of the line are :
•  13,13,13
•   1 ,1, 1
•  1,2,3
•  13,13,13
Solution
Given lines can be rewritten as (x-1/3)/1=(y-1/3)/2=(z-1)/3 This shows that DR’s of given equation are (1, 2, 3). .

Q8. If a line lies in the octant OXYZ and it makes equal angles with the axes, then :
•  l=m=n±13
•  T l=m=n=13
•  l=m=n=-13
•   l=m=n=±13
Solution
Since, it is given that line makes equal angle with the coordinate axes ∴l=m=n We know, l^2+m^2+n^2=1 ⇒3l^2=1 ⇒l^2=1/3 ⇒l=1/√3 (neglect –ve sign) .

Q9. The image of the point (5, 4, 6) in the plane x+y+2z-15=0 is :
•  (-5, -4, -6)
•  (2, 3, 2)
•  (2, 2, 3)
•  (3, 2, 2)
• Solution
The image (x,y,z) of a point (x_1,y_1,z_1) in a plane ax+by+cz+d=0 is (x-x_1)/a=(y-y_1)/b=(z-z_1)/c = (-2(ax_1+by_1+cz_1+d))/(a^2+b^2+c^2 ) Here, (x_1,y_1,z_1 )=(5,4,6) a=1, b=1,c=2,d=-15 ∴ (x-5)/1=(y-4)/1=(z-6)/2 =(-2(5+4+12-15))/(1+1+4)=-2 ⟹x=3,y=2,z=2

Q10.  The equation of the line of intersection of the planes x+2y+z=3 and 6x+8y+3z=13 can be written as
•  x-22=y+1-3=z-34
•   x+22=y-1-3=z-34
•   x-22=y+13=z-34
• x+22=y+23=z-34
Solution
Let the DR’s of a required line be a,b and c Since, the normal to the given planes x+2y+z=3 and 6x+8y+3z=13 are perpendicular to the line. ∴a+2b+c=0 and 6a+8b+3c=0 ⟹a/(6-8)=b/(6-3)=c/(8-12) ⟹a/2=b/(-3)=c/4 Hence, option ( c) is the required solution. .

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