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.**The vector equation of a plane which contains the line $\overrightarrow{r}$=2î+Î»(ĵ-$\hat{k}$) and perpendicular to the plane $\overrightarrow{r}$∙(î+$\hat{k}$)=3, is

Solution

(a) Since the required plane contains the line $\overrightarrow{r}$=2î+Î»(ĵ-$\hat{k}$) and is perpendicular to the plane $\overrightarrow{r}$∙(î+$\hat{k}$)=3.

Therefore, it passes through the point $\overrightarrow{a}$=2î̂ and is parallel to the vectors $\overrightarrow{b}$=ĵ-$\hat{k}$and $\overrightarrow{c}$=î+k ̂.

Hence, it is perpendicular to the vector $\overrightarrow{n}$=$\overrightarrow{b}$×$\overrightarrow{c}$=(ĵ-$\hat{k}$)×(î+$\hat{k}$)=î-ĵ-k ̂

Therefore, the equation of the required plane is ($\overrightarrow{r}$-$\overrightarrow{a}$ )∙$\overrightarrow{n}$=0 ⇒($\overrightarrow{r}$-2î̂ )∙(î-ĵ-$\hat{k}$)=0 ⇒$\overrightarrow{r}$∙(î-ĵ-$\hat{k}$)=2

(a) Since the required plane contains the line $\overrightarrow{r}$=2î+Î»(ĵ-$\hat{k}$) and is perpendicular to the plane $\overrightarrow{r}$∙(î+$\hat{k}$)=3.

Therefore, it passes through the point $\overrightarrow{a}$=2î̂ and is parallel to the vectors $\overrightarrow{b}$=ĵ-$\hat{k}$and $\overrightarrow{c}$=î+k ̂.

Hence, it is perpendicular to the vector $\overrightarrow{n}$=$\overrightarrow{b}$×$\overrightarrow{c}$=(ĵ-$\hat{k}$)×(î+$\hat{k}$)=î-ĵ-k ̂

Therefore, the equation of the required plane is ($\overrightarrow{r}$-$\overrightarrow{a}$ )∙$\overrightarrow{n}$=0 ⇒($\overrightarrow{r}$-2î̂ )∙(î-ĵ-$\hat{k}$)=0 ⇒$\overrightarrow{r}$∙(î-ĵ-$\hat{k}$)=2

**Q2.**The line passing the points 6$\overrightarrow{a}$-4$\overrightarrow{b}$+4$\overrightarrow{c}$,-4$\overrightarrow{c}$ and the line joining the points -$\overrightarrow{a}$-2$\overrightarrow{b}$-3$\overrightarrow{c}$,$\overrightarrow{a}$+2$\overrightarrow{b}$-5$\overrightarrow{c}$ intersect at:

Solution

(d) The equations of the lines joining 6$\overrightarrow{a}$-4$\overrightarrow{b}$+4$\overrightarrow{c}$,-4$\overrightarrow{c}$ and -$\overrightarrow{a}$-2$\overrightarrow{b}$-3$\overrightarrow{c}$,$\overrightarrow{a}$+2$\overrightarrow{b}$-5$\overrightarrow{c}$ are respectively

$\overrightarrow{r}$=6$\overrightarrow{a}$-4$\overrightarrow{b}$+4$\overrightarrow{c}$+m(-6$\overrightarrow{a}$-4$\overrightarrow{b}$-8$\overrightarrow{c}$ ) …(i) and, $\overrightarrow{r}$=-$\overrightarrow{a}$-2$\overrightarrow{b}$-3$\overrightarrow{c}$+n(2$\overrightarrow{a}$+4$\overrightarrow{b}$-2$\overrightarrow{c}$ ) …(ii)

For the point of intersection, the equations (i) and (ii) should give the same value of $\overrightarrow{r}$

Hence, equating the coeff. of vectors $\overrightarrow{a}$,$\overrightarrow{b}$ and $\overrightarrow{c}$ in the two expressions for $\overrightarrow{r}$, we get

6m+2n=7,2m-2n=1 and 8m-2n=7

Solving first two equations, we get m=1,n=1/2

These values of m and n also satisfy the third equation

Hence, the lines intersect

Putting the value of m in (i), we obtain that the position vector of the point of intersection as -4$\overrightarrow{c}$

(d) The equations of the lines joining 6$\overrightarrow{a}$-4$\overrightarrow{b}$+4$\overrightarrow{c}$,-4$\overrightarrow{c}$ and -$\overrightarrow{a}$-2$\overrightarrow{b}$-3$\overrightarrow{c}$,$\overrightarrow{a}$+2$\overrightarrow{b}$-5$\overrightarrow{c}$ are respectively

$\overrightarrow{r}$=6$\overrightarrow{a}$-4$\overrightarrow{b}$+4$\overrightarrow{c}$+m(-6$\overrightarrow{a}$-4$\overrightarrow{b}$-8$\overrightarrow{c}$ ) …(i) and, $\overrightarrow{r}$=-$\overrightarrow{a}$-2$\overrightarrow{b}$-3$\overrightarrow{c}$+n(2$\overrightarrow{a}$+4$\overrightarrow{b}$-2$\overrightarrow{c}$ ) …(ii)

For the point of intersection, the equations (i) and (ii) should give the same value of $\overrightarrow{r}$

Hence, equating the coeff. of vectors $\overrightarrow{a}$,$\overrightarrow{b}$ and $\overrightarrow{c}$ in the two expressions for $\overrightarrow{r}$, we get

6m+2n=7,2m-2n=1 and 8m-2n=7

Solving first two equations, we get m=1,n=1/2

These values of m and n also satisfy the third equation

Hence, the lines intersect

Putting the value of m in (i), we obtain that the position vector of the point of intersection as -4$\overrightarrow{c}$

**Q3.**The point of intersection of the lines $$$$\frac{(x-5)} {3}=\frac{(y-7)}{(-1)}=\frac{(z+2)}{1}$$ $$\frac{(x+3)}{(-36)}=\frac{(y-3)}{2}=\frac{(z-6)}{4}$$ is:

Solution

(b) Given equation of lines are

(x-5)/3=(y-7)/(-1)=(z+2)/1=k [say]…..(i)

and (x+3)/(-36)=(y-3)/2=(z-6)/4 (ii)

Any point on the line (i) is P(3k+5,-k+7,k-2)

This point is satisfied the Eq. (ii),

∴(3k+5+3)/(-36)=(-k+7-3)/2=(k-2-6)/4

⟹(3k+8)/(-36)=(-k+4)/2=(k-8)/4

⟹3k+8=18k-72⟹k=16/3

∴P(16+5,-16/3+7,16/3-2)

ie,P(21,5/3,10/3)

(b) Given equation of lines are

(x-5)/3=(y-7)/(-1)=(z+2)/1=k [say]…..(i)

and (x+3)/(-36)=(y-3)/2=(z-6)/4 (ii)

Any point on the line (i) is P(3k+5,-k+7,k-2)

This point is satisfied the Eq. (ii),

∴(3k+5+3)/(-36)=(-k+7-3)/2=(k-2-6)/4

⟹(3k+8)/(-36)=(-k+4)/2=(k-8)/4

⟹3k+8=18k-72⟹k=16/3

∴P(16+5,-16/3+7,16/3-2)

ie,P(21,5/3,10/3)

**Q4.**Distance between two parallel planes 4x+2y+4z+5=0 and 2x+y+2z-8=0 is :

Solution

(a) Given, planes are 2x+y+2z+5/2=0

and 2x+y+2z-8=0

∴Distance=|(5/2-(-8))/√(2^2+1^2+2^2 )|=7/2

(a) Given, planes are 2x+y+2z+5/2=0

and 2x+y+2z-8=0

∴Distance=|(5/2-(-8))/√(2^2+1^2+2^2 )|=7/2

**Q5.**The equation of the plane through the point (2, 3, 1) and (4,-5,3) and parallel to x-axis is:

Solution

(b) The equation of a plane passing through (2, 3, 1) is

a(x-2)+b(y-3)+c(z-1)=0 … (i)

It passes through (4,-5,3) and is parallel to x-axis

2a-8b+2c=0

and, a×1+b×0+c×0=0

∴a/0=b/2=c/8⇒a/0=b/1=c/4

Substituting the values of a,b,c in (i), we get y+4z=7 as the equation of the required plane

(b) The equation of a plane passing through (2, 3, 1) is

a(x-2)+b(y-3)+c(z-1)=0 … (i)

It passes through (4,-5,3) and is parallel to x-axis

2a-8b+2c=0

and, a×1+b×0+c×0=0

∴a/0=b/2=c/8⇒a/0=b/1=c/4

Substituting the values of a,b,c in (i), we get y+4z=7 as the equation of the required plane

**Q6.**The points A(-1,3,0),B(2,2,1) and C(1,1,3) determine a plane. The distance from the plane to the point D(5,7,8) is:

Solution

(a) Equation of plane passing through (-1,3,0) is

A(x+1)+B(y-3)+C(z-0)=0 ….(i)

Also, plane (i) is passing through the points (2, 2, 1) and (1,1,3)

3A-B+C=0 …(ii)

And 2A-2B+3C=0 …(iii)

On solving Eqs. (i) and (iii), we get

A/(-3+2)=B/(2-9)=C/(-6+2)

∴A:B:C=-1:-7:-4

⇒A:B:C=1:7:4

From Eq. (i), 1(x+1)+7(y-3)+4(z)=0

⇒x+7y+4z-20=0

∴ Distance from the plane to the point (5, 7, 8)

=(1×5+7×7+4×8-20)/√(1^2+7^2+4^2 )

=(5+49+32-20)/√66=66/√66=√66

(a) Equation of plane passing through (-1,3,0) is

A(x+1)+B(y-3)+C(z-0)=0 ….(i)

Also, plane (i) is passing through the points (2, 2, 1) and (1,1,3)

3A-B+C=0 …(ii)

And 2A-2B+3C=0 …(iii)

On solving Eqs. (i) and (iii), we get

A/(-3+2)=B/(2-9)=C/(-6+2)

∴A:B:C=-1:-7:-4

⇒A:B:C=1:7:4

From Eq. (i), 1(x+1)+7(y-3)+4(z)=0

⇒x+7y+4z-20=0

∴ Distance from the plane to the point (5, 7, 8)

=(1×5+7×7+4×8-20)/√(1^2+7^2+4^2 )

=(5+49+32-20)/√66=66/√66=√66

**Q7.**The angle between $$\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) and the plane 3x+2y-3z=4,is :

Solution

(b) sin〖Î¸=(a_1 a_2+b_1 b_2+c_1 c_2)/(√(a_1^2+b_1^2+c_1^2 ) √(a_2^2+b_2^2+c_2^2 ))〗

=(2×3+3×2-4×3)/(√(2^2+3^2+4^2 ) √(〖(3)〗^2+〖(2)〗^2+〖(-3)〗^2 ))

=(6+6-12)/(√(4+9+16) √(9+4+9))=0

⟹ Î¸=0°

(b) sin〖Î¸=(a_1 a_2+b_1 b_2+c_1 c_2)/(√(a_1^2+b_1^2+c_1^2 ) √(a_2^2+b_2^2+c_2^2 ))〗

=(2×3+3×2-4×3)/(√(2^2+3^2+4^2 ) √(〖(3)〗^2+〖(2)〗^2+〖(-3)〗^2 ))

=(6+6-12)/(√(4+9+16) √(9+4+9))=0

⟹ Î¸=0°

**Q8.**A point on XOZ- plane divides the join of (5,-3,-2) and (1,2,-2) at:

Solution

(a) Let the point P(x,y,z) divides the line joining the points A and B in the ratio m:1.

Since, point P is on XOZ-plane

∴y coordinate = 0

⟹ (2m-3)/(m+1)=0 ⟹m=3/2

Now,x=(3+2×5)/(3+2)=13/5

and z=(3×(-2)+2×(-2))/5=-2

∴Required points is (13/5,0,-2)

(a) Let the point P(x,y,z) divides the line joining the points A and B in the ratio m:1.

Since, point P is on XOZ-plane

∴y coordinate = 0

⟹ (2m-3)/(m+1)=0 ⟹m=3/2

Now,x=(3+2×5)/(3+2)=13/5

and z=(3×(-2)+2×(-2))/5=-2

∴Required points is (13/5,0,-2)

**Q9.**The plane passing through the point (5, 1, 2) perpendicular to the line 2(x-2)=y-4=z-5 will

meet the line in the point:

Solution

(a) Equation of the plane through (5, 1, 2) is

a(x-5)+b(y-1)+c(z-2)=0 …(i)

Given plane (i) is perpendicular to the line

(x-2)/(1/2)=(y-4)/1=(z-5)/1

∴ Equation of normal of Eq. (i) and straight line (ii) are parallel

ie,a/(1/2)=b/1=c/1=k (say)

∴a=k/2,b=k,c=k

From Eq. (i),

k/2 (x-5)+k(y-1)+k(z-2)=0

Or x+2y+2z=11

Any point on Eq. (ii) is (2+Î»/2,4+Î»,5+Î»)

Which lies on Eq. (iii), then Î»=-2

∴ Required point is (1, 2, 3)

(a) Equation of the plane through (5, 1, 2) is

a(x-5)+b(y-1)+c(z-2)=0 …(i)

Given plane (i) is perpendicular to the line

(x-2)/(1/2)=(y-4)/1=(z-5)/1

∴ Equation of normal of Eq. (i) and straight line (ii) are parallel

ie,a/(1/2)=b/1=c/1=k (say)

∴a=k/2,b=k,c=k

From Eq. (i),

k/2 (x-5)+k(y-1)+k(z-2)=0

Or x+2y+2z=11

Any point on Eq. (ii) is (2+Î»/2,4+Î»,5+Î»)

Which lies on Eq. (iii), then Î»=-2

∴ Required point is (1, 2, 3)

**Q10.**If $\overrightarrow{r}$ is a vector of magnitude 21 and has direction ratios proportional to 2,-3,6, then $\overrightarrow{r}$ is equal to:

Solution

(a) It is given that the direction ratios of $\overrightarrow{r}$ are proportional to 2,-3,6.

Therefore, its direction cosines are l=2/7,m=(-3)/7,n=6/7

∴$\overrightarrow{r}$=|$\overrightarrow{r}$ |(l î+m( ĵ) ̂+n $\hat{k}$) ⇒$\overrightarrow{r}$=21(2/7 î-3/7 ĵ+6/7 $\hat{k}$)=6î-9ĵ+18 $\hat{k}$

(a) It is given that the direction ratios of $\overrightarrow{r}$ are proportional to 2,-3,6.

Therefore, its direction cosines are l=2/7,m=(-3)/7,n=6/7

∴$\overrightarrow{r}$=|$\overrightarrow{r}$ |(l î+m( ĵ) ̂+n $\hat{k}$) ⇒$\overrightarrow{r}$=21(2/7 î-3/7 ĵ+6/7 $\hat{k}$)=6î-9ĵ+18 $\hat{k}$