## Three Dimensional Geometry Quiz

Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

Q1. Let a =i ̂+j ̂ and b =2i ̂ - k ̂, then the point of intersection of the lines r×a =b×a and r×b =a×b is
•  (3,-1,1)
•  (3,1,-1)
•  (-3,1,1)
•  (-3,-1,-1)
Solution
1 (b) Let r ×a =b ×a ⇒(r ⃗-b ⃗ )×a ⃗=0 ⃗ ⇒ r ⃗=b ⃗+ta ⃗ Similarly, other line r =a +kb , where t and k are scalars Now a +kb =b +ta ⇒ t=1,k=1 (equating the coefficients of a ⃗ and b ⃗) ∴ r ⃗=a ⃗+b ⃗=i ̂+j ̂+2i ̂-k ̂=3i ̂+j ̂-k ̂ i.e., (3,1,-1)

Q2.The distance between the line: r =2i ̂-2j ̂+3k ̂+Î»(i ̂-j ̂+4k ̂) and the plane r ∙(i ̂+5j ̂+k ̂ )=5 is
•  10/(3√3)
•  10/9
•  10/3
•  3/10
Solution
(a) It is obvious that the given line and plane are parallel Given point on the line is A(2,-2,3) B(0,0,5) is a point on the plane ∴ (AB) ⃗=(2-0) i ̂+(-2-0) j ̂+(3-5) k ̂ Then distance of B from the plane = projection of (AB) ⃗ on vector i ̂+5j ̂+k ̂ p=|((2i ̂-2j ̂-2k ̂ )∙(i ̂+5j ̂+k ̂))/√(1+25+1)| =|(2-10-2)/√27|=10/(3√3)

Q3.  The projection of point P(p ⃗) on the plane r ⃗∙n ⃗=q is (s ⃗), then
•   a) s =((q-p ∙n )n )/|n |^2
•  b) s =p +((q-p ∙n )n )/|n |^2
•  s =p -((p ∙n ∙)n )/|n |^2
•  s =p -((q-p ∙n )n )/|n |^2
Solution

Q4. The intercepts made on the axes by the plane which bisects the line joining the points (1,2, 3) and (-3,4,5) at right angles are
•  (-9/2,9,9)
•  (9/2,9,9)
•  (9,-9/2,9)
•  (9,9/2,9)
Solution
(a) Direction ratios of the line joining points P(1,2,3) and Q(-3,4,5) are -4,2,2 which are direction ratios of the normal to the plane Then, equation of plane is -4x+2y+2z=k Also this plane passes through the midpoint of PQ(-1,3,4) ⇒ -4(-1)+2(3)+2(4)=k ⇒k=18 ⇒ Equation of plane is 2x-y-z=-9 Then, intercepts are (-9/2),9 and 9

Q5.The plane r ∙n =q will contain the line r =a +Î»b , if
•   b ∙n ≠0,a ∙n ⃗≠q
•  b ∙n =0,a ∙n ≠q
•   b ∙n =0,a ∙n =q
•   b ∙n ≠0,a ∙n =q
Solution
(c) We must have b ∙n =0 and a ∙n =q

Q6. The vector equation of the plane passing through the origin and the line of intersection of the planes r ⃗∙a ⃗=Î» and r ∙b =Î¼ is
•  r ∙(Î»a -Î¼b )=0
•  r ∙(Î»b -Î¼a )=0
• r ∙(Î»a +Î¼b )=0
•  r ∙(Î»b +Î¼a )=0
Solution
(b) The equation of a plane through the line of intersection of the planes r .a =Î» and r .b =Î¼ is (r .a -Î»)+k(r .b -Î¼)=0 or r .(a +kb )=Î»+kÎ¼ (i) This passes through the origin, therefore 0 (a +kb )=Î»+Î¼k ⇒k=(-Î»)/Î¼ Putting the value of k in (i). we get the equation of the required plane as r .(Î¼a -Î»b )=0 ⇒ r .(Î»b -Î¼a )=0

Q7.The line (x-2)/3=(y+1)/2=(z-1)/(-1) intersects the curve xy=c^2,z=0 if c is equal to
•  ±1
•  ±1/3
•  ±√5
•  None of these
Solution
(c) We have z=0 for the point, where the line intersects the curve Therefore, (x-2)/3=(y+1)/2=(0-1)/(-1) ⇒(x-2)/3=1 and (y+1)/2=1 ⇒x=5 and y=1 Putting these values in xy=c^2, we get 5=c^2⇒ c=±√5

Q8. L_1 and L_2 are two lines whose vector equations are L_1:r ⃗=Î»((cos⁡Î¸+√3) i ̂+(√2 sin⁡Î¸ ) j ̂+(cos⁡Î¸-√3) k ̂ ) L_2:r ⃗=Î¼(ai ̂+bj ̂+ck ̂ ), where Î» and Î¼ are scalars and Î± is the acute angle between L_1 and L_2. If the angle ‘Î±’ is independent of Î¸, then the value of ‘Î±’ is
•  Ï€/6
•  TÏ€/4
•  Ï€/3
•  Ï€/2
Solution
(a) Both the lines pass through origin. Line L_1 is parallel to the vector V _1 V _1=(cos⁡Î¸+√3) i ̂+(√2 sin⁡Î¸ ) j ̂+(cos⁡Î¸-√3)k ̂ and L_2 is parallel to the vector V _2 V _2=ai ̂+bj ̂+ck ̂ ∴ cos⁡Î±=(V _1.V _2)/(|V _1 ||V_2 |) =(a(cos⁡Î¸+√3)+(b√2) sin⁡Î¸@+c(cos⁡Î¸-√3))/(√(a^2+b^2+c^2 )@ √((cos⁡Î¸+√3)^2+2 sin^2⁡Î¸+(cos⁡Î¸-√3)^2 )) =((a+c) cos⁡Î¸+b√2 sin⁡Î¸+(a-c)√3)/(√(a^2+b^2+c^2 ) √(2+6)) For cos⁡Î± to be independent of Î¸, we get a+c=0 and b=0 ∴ cos⁡Î±=(2a√3)/(a√2 2√2)=√3/2 ⇒ Î±=Ï€/6

Q9. Distance of the point P(p ) from the line r =a +Î»b is
•  a) |(a -p )+(((p -a )∙b )b )/|b |^2 |
•  b) |(b -p )+(((p -a )∙b )b )/|b |^2 |
•  c) |(a -p )+(((p -b )∙b )b )/|b |^2 |
•  None of these
Solution
(c) Let Q(q ) be the foot of altitude drawn from P(p ) to the line r =a +Î»b , ⇒ (q ⃗-p ⃗ )∙b ⃗=0 and q ⃗=a ⃗+Î»b ⇒(a ⃗+Î»b -p )∙b =0 ⇒(a ⃗-p ⃗ )∙b ⃗+Î»|b |^2=0 ⇒ Î»= ((p -a )∙b )/|b |^2 ⇒ q ⃗-p ⃗=a ⃗+(((p ⃗-a ⃗ )∙b ⃗)b ⃗)/|b ⃗ |^2 -p ⃗ ⇒|q ⃗-p ⃗ |=|(a ⃗-p ⃗ )+(((p ⃗-a ⃗ )∙b ⃗)b ⃗)/|b ⃗ |^2 |

Q10. The coordinates of the foot of the perpendicular drawn from the origin to the line joining the points (-9,4,5) and (10,0,-1) will be
•  (-3,2,1)
•  (1, 2, 2)
•  (4, 5, 3)
• None of these
Solution
10 (d) Let AD be the perpendicular and D be the foot of the perpendicular which divides BC in the ratio Î»:1, then D((10Î»-9)/(Î»+1),4/(Î»+1),(-Î»+5)/(Î»+1)) The direction ratios of AD are (10Î»-9)/(Î»+1),4/(Î»+1) and (–Î»+5)/(Î»+1) and direction ratios of BC are 19,-4 and -6 Since AD⊥BC, we get 19((10Î»-9)/(Î»+1))-4(4/(Î»+1))-6((-Î»+5)/(Î»+1))=0 ⇒ Î»=31/28 Hence, on putting the value of Î» in (i), we get required foot of the perpendicular, i.e., (58/59,112/59,109/59)

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