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.**The length of the perpendicular drawn from (1, 2, 3) to the line (x-6)/3=(y-7)/2=(z–7)/(-2) is

Solution

(d) Let P be the point (1,2,3) and PN be the length of the perpendicular from P on the given line Coordinates of point N are (3Î»+6,2Î»+7,-2Î»+7) Now PN is perpendicular to the given line or vector 3i ̂+2j ̂-2k ̂ ⇒3(3Î»+6-1)+2(2Î»+7-2)-2(-2Î»+7-3)=0 ⇒ Î»=-1 Then, point N is (3, 5, 9) ⇒PN=7

(d) Let P be the point (1,2,3) and PN be the length of the perpendicular from P on the given line Coordinates of point N are (3Î»+6,2Î»+7,-2Î»+7) Now PN is perpendicular to the given line or vector 3i ̂+2j ̂-2k ̂ ⇒3(3Î»+6-1)+2(2Î»+7-2)-2(-2Î»+7-3)=0 ⇒ Î»=-1 Then, point N is (3, 5, 9) ⇒PN=7

**Q2.**If angle Î¸ between the line (x+1)/1=(y-1)/2=(z-2)/2 and the plane 2x-y+√Î» z+4=0 is such that sinÎ¸=1/3, the value of Î» is

Solution

(b) The line is (x+1)/1=(y-1)/2=(z-2)/2 and the plane is 2x-y+√Î» z+4=0 If Î¸ be the angle between the line and the plane, then 90°-Î¸ is the angle between the line and normal to the plane ⇒cos(90°-Î¸)=((1)(2)+(2)(-1)+(2)(√Î»))/(√(1+4+4) √(4+1+Î»)) ⇒sinÎ¸=(2-2+2√Î»)/(3√(5+Î»)) ⇒1/3=(2√Î»)/(3√(5+Î»)) ⇒√(5+Î»)=2√Î» ⇒5+Î»=4Î» ⇒3Î»=5 ⇒ Î»=5/3

(b) The line is (x+1)/1=(y-1)/2=(z-2)/2 and the plane is 2x-y+√Î» z+4=0 If Î¸ be the angle between the line and the plane, then 90°-Î¸ is the angle between the line and normal to the plane ⇒cos(90°-Î¸)=((1)(2)+(2)(-1)+(2)(√Î»))/(√(1+4+4) √(4+1+Î»)) ⇒sinÎ¸=(2-2+2√Î»)/(3√(5+Î»)) ⇒1/3=(2√Î»)/(3√(5+Î»)) ⇒√(5+Î»)=2√Î» ⇒5+Î»=4Î» ⇒3Î»=5 ⇒ Î»=5/3

**Q3.**Let L be the line of intersection of the planes 2x+3y+z=1 and x+3y+2z=2. If L makes an angle Î± with the positive x-axis, then cosÎ± equals

**Q6.**The point of intersection of the lines (x-5)/3=(y-7)/(-1)=(z+2)/1 and (x+3)/(-36)=(y-3)/2=(z-6)/4 is

**Q7.**What is the equation of the plane which passes through the z-axis and is perpendicular to the line (x-a)/cosÎ¸ =(y+2)/sinÎ¸ =(z-3)/0?

**Q8.**The line (x+6)/5=(y+10)/3=(z+14)/8 is the hypotenuse of an isosceles right angled triangle whose opposite vertex is (7,2,4). Then which of the following is not the side of the triangle?

Solution

(c) Given one vertex A(7,2,4) and line (x+6)/5=(y+10)/3=(z+14)/8 General point on above line B≡(5Î»-6,3Î»-10,8Î»-14) Direction ratios of line AB are <5Î»-13,3Î»-12,8Î»-18> Direction ratios of line BC are <5,3,8> Since angle between AB nad BC is Ï€/4 cos〖Ï€/4〗=((5Î»-3)5+3(3Î»-12)+8(8Î»-18))/(√(5^2+3^2+8^2 )∙√((5Î»-13)^2+(3Î»-12)^2+(8Î»-18)^2 )) Squaring and solving, we have Î»=3,2 Hence equation of lines are (x-7)/2=(y-2)/(-3)=(z-4)/6 and (x-7)/3=(y-2)/6=(z-4)/2

(c) Given one vertex A(7,2,4) and line (x+6)/5=(y+10)/3=(z+14)/8 General point on above line B≡(5Î»-6,3Î»-10,8Î»-14) Direction ratios of line AB are <5Î»-13,3Î»-12,8Î»-18> Direction ratios of line BC are <5,3,8> Since angle between AB nad BC is Ï€/4 cos〖Ï€/4〗=((5Î»-3)5+3(3Î»-12)+8(8Î»-18))/(√(5^2+3^2+8^2 )∙√((5Î»-13)^2+(3Î»-12)^2+(8Î»-18)^2 )) Squaring and solving, we have Î»=3,2 Hence equation of lines are (x-7)/2=(y-2)/(-3)=(z-4)/6 and (x-7)/3=(y-2)/6=(z-4)/2

**Q9.**The distance of point A(-2,3,1) from the line PQ through P(-3,5,2), which makes equal angles with the axes is

**Q10.**From the point P(a,b,c), let perpendicular PL and PM be drawn to YOZ and ZOX planes, respectively. Then the equation of the plane OLM is