As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced..

**Q1.**If the angle between the line $$ \(\frac{(x+1)}{1}=\frac{(y-1)}{2}=\frac{(z-2)}{2}\) and the plane 2x-y+√Î» z+4=0 is such that sinÎ¸=1/3 Then, value of Î» is:

**Q2.**Let the line$$\(\frac{(x-2)}{3}=\frac{(y-1)}{-5}=\frac{(z+2)}{2}\) lies in the plane x+3y-Î±z+Î²=0 Then (Î±,Î²) equals:

Solution

(b)

DR’s of given line are (3,-5,2)

DR’s of normal to the plane =(1,3,-Î±)

∴ Line is perpendicular to the normal

⟹3(1)-5(3)+2(-Î±)=0

⟹3-15-2Î±=0

⟹2Î±=-12

⟹ Î±=-6

Also point (2,1,-2) lies on the plane

2+3+6(-2)+Î²=0

⟹ Î²=7

∴(Î±,Î²)=(-6,7)

(b)

DR’s of given line are (3,-5,2)

DR’s of normal to the plane =(1,3,-Î±)

∴ Line is perpendicular to the normal

⟹3(1)-5(3)+2(-Î±)=0

⟹3-15-2Î±=0

⟹2Î±=-12

⟹ Î±=-6

Also point (2,1,-2) lies on the plane

2+3+6(-2)+Î²=0

⟹ Î²=7

∴(Î±,Î²)=(-6,7)

**Q3.**The distance between the line $\overrightarrow{\mathrm{r.}}$=2$\hat{i}$-2$\hat{j}$ +3$\hat{k}$ +Î»($\hat{i}$ -$\hat{j}$ +4$\hat{k}$) and the plane $\overrightarrow{\mathrm{r.}}$($\hat{i}$+5$\hat{j}$ +$\hat{k}$)=5, is:

**Q4.**In a three dimensional xyz-space, the equation x

^{2}-5x+6=0 represents:

Solution

(b)

Given equation is x

⇒(x-2)(x-3)=0

⇒(x-2)=0 or (x-3)=0

Which represents a plane

(b)

Given equation is x

^{2}-5x+6=0⇒(x-2)(x-3)=0

⇒(x-2)=0 or (x-3)=0

Which represents a plane

**Q5.**The direction ratios of a normal to the plane passing through (0, 0, 1), (0, 1, 2) and (1, 2, 3) are proportional to:

**Q6.**The image of the point (1, 2, 3) in lie $$ \(\frac{x}{2}=\frac{(y-1)}{3}=\frac{(z-1)}{3} \)is

**Q7.**The equation of the plane passing through (1, 1, 1) and (1,-1,-1) and perpendicular to 2x-y+z+5=0 is :

**Q8.**If a sphere of radius r passes through the origin, then the extremities of the diameter parallel to x-axis lie on each of the spheres:

Solution

(a)

Let (u,v,w) be the centre of the sphere with radius r. Since, it passes through the origin

∴u

$$\(\frac{(x-u)}{1}=\frac{(y-v)}{0}=\frac{(z-w)}{0} \) …(i)

As it passes through u,v,w and direction ratios of x- axis are 1, 0, 0

The extremities of diameter are the points on Eq. (i) at a distance r from the centre (u,v,w)

∴ The required extremities are P(r+u,v,w) and Q(-r+u,v,w)

P lies on the sphere x

(r+u)

Because u

and similarly Q lies on the sphere x

(a)

Let (u,v,w) be the centre of the sphere with radius r. Since, it passes through the origin

∴u

^{2}+v^{2}+w^{2}=r^{2}. Equation of the diameter parallel to x-axis is$$\(\frac{(x-u)}{1}=\frac{(y-v)}{0}=\frac{(z-w)}{0} \) …(i)

As it passes through u,v,w and direction ratios of x- axis are 1, 0, 0

The extremities of diameter are the points on Eq. (i) at a distance r from the centre (u,v,w)

∴ The required extremities are P(r+u,v,w) and Q(-r+u,v,w)

P lies on the sphere x

^{2}+y^{2}+z^{2}-2rx=0 as(r+u)

^{2}+v^{2}+w^{2}-2r(r+u)=0Because u

^{2}+v^{2}+w^{2}=r^{2}and similarly Q lies on the sphere x

^{2}+y^{2}+z^{2}+2rx=0**Q9.**A And B are two give points. Let C divides AB internally and D divides AB externally in the same ratio. Then AC,AB,AD are in:

Solution

Not Available

Not Available

**Q10.**The length of the perpendicular from the origin to the plane passing through the point $\overrightarrow{a}$ and containing the line $\overrightarrow{r}$ =$\overrightarrow{b}$ +Î» $\overrightarrow{c}$ is: