Plane | Part-1 | Ex-5A

Plane
MCQ related solutions

Choose the correct option :

  1. The planes ~bx-ay=n,~cy-bz=l,~az-cx=m~ intersect in a line if-

~(i)~al+bm+cn=1,~~(ii)~ al-bm-cn=0,~~(iii)~al+bm+cn=0~(iv)~ none of these.

Solution.

The equation of any plane passing through the line of intersection of two planes can be written as ~ P_1+\lambda P_2=0,~(\lambda \neq 0).

\text{Let}~~P_1=bx-ay-n=0,~~P_2=cy-bz-l=0,~~P_3=az-cx-m=0\rightarrow(1)

\text{Now,}~P_1+\lambda P_2=0 \\ \text{or,}~~bx-ay-n+\lambda(cy-bz-l)=0 \\ \text{or,}~~ bx+y(\lambda c-a)-n-\lambda bz-\lambda l=0\rightarrow(2)

Comparing ~(1)~ and ~(2)~ we get,

~b=-c,~ \lambda c-a=0,~~ a=-\lambda b,\\~~ n+\lambda l=m \rightarrow(3)

~~\text{Now,}~~\lambda c-a=0 \Rightarrow \lambda =\frac ac.

From (3) we get, 

~ n+\frac ac \cdot l=m~~(\because \lambda=\frac ac) \\ \text{or,}~~ nc+al=mc \\ \text{or,}~~ nc+al=-mb~~(\because b=-c) \\ \text{or,}~~ al+mb+nc=0.

So, option (c) is correct.

2. The point in which the line ~\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}~meets the plane ~x-2y+z=20~ is-

~(a)~(8,7,26)~~(ii)~(-8,7,26)~~(iii)~(8,-7,26)~~(iv)~(8,7,-26)

Solution.

~\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t \\ \therefore~~x-2=3t \Rightarrow x=3t+2, \\~y+1=4t \Rightarrow y=4t-1,~\\~~z-2=12t \Rightarrow z=12t+2.

Let the straight line meets the plane at the point ~(3t+2,4t-1,12t+2).

~\text{So,}~x-2y+z=20 \\ \text{or,}~~ 3t+2-2(4t-1)+12t+2=20\\ \text{or,}~~ 15t+4-8t+2=20 \\ \text{or,}~~ 7t+6=20 \\ \text{or,}~~ t=\frac{20-6}{7}=2.

So, the required point is ~(3 \times 2+2, 4 \times 2-1, 12 \times 2+2)=(8,7,26)

So, option (a) is correct.

3. The coordinates of the point where the line joining the points (2,-3,1) and (3,4,-5) meets the xy plane are-

(a) \left(-\frac{13}{6},\frac{11}{6},0\right)  (b) \left(\frac{13}{6},-\frac{11}{6},0\right)  (c) \left(\frac{13}{6},\frac{11}{6},0\right) (d) none of these

Solution.

Let ~A=(2,-3,1),~B=(3,4,-5).

\therefore  \vec{a} =2\hat{i}-3\hat{j}+\hat{k},~~\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}.

The equation of the straight line passing through ~\vec{a}~ and ~\vec{b}~ is given by 

~\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a}) \\ \text{or,}~~ \vec{r}=2\hat{i}-3\hat{j}+\hat{k}+\lambda [(3\hat{i}+4\hat{j}-5\hat{k})-(2\hat{i}-3\hat{j}+\hat{k})] \\ \text{or,}~~ \vec{r}=(2+\lambda)\hat{i}+(-3+7\lambda)\hat{j}+(1-6\lambda)\hat{k}\rightarrow(1)

Let the straight line (1) intersects the plane xy at the point ~(x,y,0).

\therefore~~\vec{r}=x\hat{i}+y\hat{j}+0 \hat{k}\rightarrow(2)

So, from (1) and (2) we get,

x=2+\lambda,~y=-3+7\lambda,~1-6\lambda=0 \Rightarrow  \lambda=\frac 16.

\therefore~x=2+\frac 16=\frac{13}{6},~y=-3+ 7 \times \frac 16=-\frac{11}{6}.

So, the point is ~\left(\frac{13}{6},-\frac{11}{6},0\right).

Hence, option (b) is correct.


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4. The line joining the points (1,1,2) and (3,-2,1) meets the plane ~3x+2y+z=6 is-

(a) (-3,-2,-1)  (b) (3,-2,1) (c) (-3,2,1) (d) (3,2,1)

Solution.

The equation of the straight line passing through the points (1,1,2) and (3,-2,1) is given by 

~\frac{x-1}{3-1}=\frac{y-1}{-2-1}=\frac{z-2}{1-2} \\ \text{or,}~~ \frac{x-1}{2}=\frac{y-1}{-3}=\frac{z-2}{-1}=t~~(\text{say})

\therefore~ x-1=2t \Rightarrow x=2t+1,\\~~y-1=-3t \Rightarrow y=-3t+1 \\ ~~z-2=-t \Rightarrow z=-t+2.

So, the point (2t+1,-3t+1,-t+2) lies on the plane.

\therefore~ 3x+2y+z=6 \\ \text{or,}~~ 3(2t+1)+2(-3t+1)+(-t+2)=6 \\ \text{or,}~~ 6t+3-6t+2-t+2=6 \\ \text{or,}~~ -t+7=6 \Rightarrow t=-1.

So, the required point is ~(2 \times 1+1,-3 \times 1+1,-1+2)=(3,-2,1).

So, option (b) is correct.

5. A plane meets the coordinate axes in A, B,C such that the centroid of the triangle ABC is the point (a, a, a). Then the equation of the plane x+y+z=p,~ where p is –

(a) 6a  (b) -3a  (c) 0 (d) 3a 

Solution. 

If (a,a,a) is the centroid of the triangle ABC, then the equation of the plane is 

x+y+z=p \Rightarrow p=a+a+a=3a.

So, option (d) is correct.

6. State which of the following statement is true ?

(a) The plane through the points ~A,B,C~ whose coordinates are ~(1,1,1),~(1,-1,1)~ and ~(-1,-3,-5)~ respectively, passes through the point ~(2,k,4)~ for all values of ~k?

(b) The equation of the plane which is parallel to the plane ~\vec{r} \cdot (2\hat{i}-3\hat{j}+5\hat{k})+7=0~ and passing through the point ~(3,4,-1)~ is ~\vec{r} \cdot (2\hat{i}-3\hat{j}+5\hat{k})+10=0.

(c) The equation of the line of intersection of planes ~3x+y+z=6~ and ~x-y+2z=5~ is ~\frac{4x-11}{3}=\frac{4y+9}{5}=\frac{z-0}{1}.

(d) The equation of the straight line ~\frac{x+3}{2}=\frac{y-4}{3}=\frac{z+5}{2}~ and  the plane ~4x-2y-z=1~ are perpendicular to each other.

Solution.

Option (a) is correct.

Explanation.

The equation of the plane through the points  ~(1,1,1),~(1,-1,1)~ and ~(-1,-3,-5)~ is 

\begin{vmatrix} x-1& y-1 &z-1  \\ 1-1& -1-1 &1-1  \\-1-1&-3-1  &-5-1  \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-1& y-1 & z-1 \\0& -2 & 0 \\ -2& -4 & -6 \\ \end{vmatrix}=0 \\ \text{or,}~~ -2\begin{vmatrix} x-1& z-1 \\ -2& -6 \\ \end{vmatrix} =0 \\ \text{or,}~~ -6(x-1)+2(z-1)=0 \\ \text{or,}~~ (z-1)-3(x-1)=0 \\ \therefore~ -3x+z+2=0\rightarrow(1)

Clearly, the point ~(2,k,4)~ satisfies the equation of the plane (1).

7. The equation of the plane passing through the point  (2,3,1) having ~5,3,2~ as the direction ratios of the normal to the plane is-

(a) 5x-3y-2z=21  (b) 5x+3y+2z=-21  (c) 5x+3y+2z=21 (d) none of these

Solution.

The equation of the plane can be written as 

a(x-x_1)+b(y-y_1)+c(z-z_1)=0\rightarrow(1) where the values of a,b,c are 5,3,2 respectively and ~(x_1,y_1,z_1)=(2,3,1).

So, by (1) we get, 

5(x-2)+3(y-3)+2(z-1)=0 \\ \text{or,}~~ 5x-10+3y-9+2z-2=0 \\ \text{or,}~~ 5x+3y+2z=21.

So, option (c) is correct.

8. The value of m for which the straight line 3x-2y+z+3=0=4x-3y+4z+1 is parallel to the plane ~2x-y+mz-2=0 is –  (a) -2 (b) 8  (c) -18 (d) 11

Solution.

The direction ratios of 3x-2y+z+3=0 are ~ 3,-2,1 and the direction ratios of ~4x-3y+4z+1=0 are ~4,-3,4.

The vector equation of the plane perpendicular to ~3x-2y+z+3=0 and ~4x-3y+4z+1=0 is given by 

\begin{vmatrix}\hat{i} &  \hat{j}&\hat{k}  \\3& -2 &1  \\ 4& -3 & 4 \\ \end{vmatrix}\\=\hat{i}(-8+3)-\hat{j}(12-4)+\hat{k}(-9+8)\\=-5\hat{i}-8\hat{j}-\hat{k}

Now, ~ -5\hat{i}-8\hat{j}-\hat{k} is perpendicular to ~2x-y+mz-2=0.

\therefore~ -5 \cdot 2-8 \cdot(-1)+m(-1)=0 \\ \text{or,}~~ -10+8-m=0 \\ \text{or,}~~ -(2+m)=0 \\ \text{or,}~~ m=-2.

So, option (a) is correct.

9. The intercept made by the plane \vec{r} \cdot \vec{n}=q~ on the x-axis is – (a) \frac{q}{\hat{i}\cdot \hat{n}}  (b) \frac{\hat{i} \cdot \hat{n}}{q} (c) \frac{\hat{i}\cdot \hat{n}}{q} (d) \frac{q}{|\vec{n}|}

Solution.

Let \vec{r}=x\hat{i}+y\hat{j}+z\hat{k},~~ \vec{n}=n_1\hat{i}+n_2\hat{j}+n_3\hat{k}

\therefore~ \vec{r} \cdot \vec{n}=q \\ \text{or,}~~ n_1x+n_2y+n_3z=q \\ \text{or,}~~ \frac{x}{q/n_1}+\frac{y}{q/n_2}+\frac{z}{q/n_3}=1

So, the required intercept is given by ~\frac{q}{n_1}=\frac{q}{\hat{i} \cdot \vec{n}}.

So, option (a) is correct.

10. A unit vector parallel to the intersection of planes ~ \vec{r} \cdot (\hat{i}-\hat{j}+\hat{k})=5~ and ~\vec{r} \cdot (2\hat{i}-3\hat{j}+\hat{k})~ is-

(a) \frac{2\hat{i}+5\hat{j}-3\hat{k}}{\sqrt{38}} (b) \frac{2\hat{i}-5\hat{j}+3\hat{k}}{\sqrt{38}} (c) \frac{2\hat{i}+5\hat{j}+3\hat{k}}{\sqrt{38}} (d) \frac{-2\hat{i}+5\hat{j}-3\hat{k}}{\sqrt{38}} 

Solution. Let the planes  ~ \vec{r} \cdot (\hat{i}-\hat{j}+\hat{k})=5~ and ~\vec{r} \cdot (2\hat{i}-3\hat{j}+\hat{k})~ be denoted by ~\vec{n_1}~ and ~n_2~ respectively.

\therefore~ \vec{n}=\vec{n_1} \times \vec{n_2}=\begin{vmatrix} \hat{i} &  \hat{j}&\hat{k}  \\ 3& -2 &1  \\ 4& -3 & 4 \\ \end{vmatrix} \\ \text{or,}~~ \vec{n}=\vec{i} (3-1)-\vec{j}(-3-2)+\vec{k}(1+2) \\ \therefore \vec{n}=2\hat{i}+5\hat{j}+3\hat{k}

Hence, the required unit vector is given by 

\frac{\vec{n}}{|\vec{n}|}=\frac{2\hat{i}+5\hat{j}+3\hat{k}}{\sqrt{2^2+5^2+3^2}}=\frac{1}{\sqrt{38}}(2\hat{i}+5\hat{j}+3\hat{k})

So, option (c) is correct.

11. The line ~\vec{r}=\vec{a}+\lambda \vec{b}~ will not meet the plane ~\vec{r} \cdot \vec{n}=\vec{q},~ if-

(a)  \vec{b}\cdot \vec{n} = 0, \vec{a} \cdot \vec{n}=q (b) \vec{b}\cdot \vec{n} \neq 0, \vec{a} \cdot \vec{n} \neq q 

(c) \vec{b}\cdot \vec{n} = 0, \vec{a} \cdot \vec{n}\neq q (d) \vec{b}\cdot \vec{n} \neq 0, \vec{a} \cdot \vec{n}=q

Solution.

\vec{r}=\vec{a}+\lambda \vec{b} \\ \therefore~ \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}+ \lambda \vec{b} \cdot \vec{n} \rightarrow(1)

If ~\vec{b} \cdot \vec{n}=0,~ then ~\vec{a} \cdot \vec{n} \neq q because the line ~\vec{r}=\vec{a} +\lambda \vec{b}~ will not meet the plane ~ \vec{r} \cdot \vec{n}=\vec{q}.

So, option (c) is correct.

12. The ratio in which the plane ~\vec{r} \cdot (\hat{i}-2\hat{j}+3\hat{k})=17~ divides the line joining the points ~-2\hat{i}+4\hat{j}+7\hat{k}~ and ~3\hat{i}-5\hat{j}+8\hat{k}~ is-

(a) ~1 : 5 (b) ~1 : 10 (c) 3 : 5 (d) 3: 10

Solution.

Let the given plane divides the line joining the points ~P(2,4,7) and ~Q(3,-5,8) in the ratio m:1

\therefore~ \vec{r} \cdot (\hat{i}-2\hat{j}+3\hat{k})=17 \\ \text{or,}~~ (x\hat{i}+y\hat{j}+z\hat{k}) \cdot (\hat{i}-2\hat{j}+3\hat{k})=17 \\ \text{or,}~~ x-2y+3z=17 \rightarrow(1)

By our supposition, ~x=\frac{3m-2}{m+1},~ y=\frac{-5m+4}{m+1},~z=\frac{8m+7}{m+1}

So, from (1) we get,

\frac{3m-2}{m+1}-2 \left(\frac{-5m+4}{m+1}\right)+3\left(\frac{8m+7}{m+1}\right)=17 \\ \text{or,}~~ 3m-2+10m-8+24m+21=17(m+1) \\ \text{or,}~~ 20m-6=0 \\ \text{or,}~~ m=\frac{6}{20}=\frac{3}{10}

Hence, option (d) is correct.

13. The lines \frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} and \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1} are coplanar if-

(a) k=1~ \text{or}~ -1  (b) k=0 ~\text{or} -3 (c) k=3 ~\text{or} -3 (d) k=0 ~\text{or} -1

Solution.

The given lines are coplanar if

\begin{vmatrix} 1-2 &  4-3 &5-4  \\ 1& 1 &-k  \\ k& 2 & 1 \\ \end{vmatrix}=\begin{vmatrix} -1 &  1&1  \\ 1& 1 &-k  \\ k& 2 & 1 \\ \end{vmatrix}=0 \\ \text{or,}~~ -1(1+2k)-1(1+k^2)+1(2-k)=0 \\ \text{or,}~~ -1-2k-1-k^2+2-k=0 \\ \text{or,}~~ -k^2-3k=0 \\ \text{or,}~~ -k(k+3)=0 \\ \text{or,}~~ k=0,-3.

So, option (b) is correct.

14. The plane , which passes through the point ~(3,2,0)~ and the line ~\frac{x-3}{1}=\frac{y-6}{5+}=\frac{z-4}{4}~ is 

(a) x-y+z=1 (b) x+y+z=5 (c) x+2y-z=1 (d) 2x-y+z=5.

Solution.

The equation of the plane passing through the point ~(3,2,0)~ is 

a(x-3)+b(y-2)+c(z-0)=0 \rightarrow(1)

Since the point (3,6,4) lies on the plane (1), so

~a(3-3)+b(6-2)+c(4-0)=0 \\ \text{or,}~~ 4b+4c=0 \Rightarrow b=-c \rightarrow(2)

Since the direction ratios of ~a, b, c~ and ~1,5,4~ are perpendicular to each other , we get

~a+5b+4c=0 \\ \text{or,}~~ a-5c+4c=0 \Rightarrow a=c \rightarrow(3)

Hence, from (1), (2) and (3) we get,

~c(x-3)-c(y-2)+c(z-0)=0 \\ \text{or,}~~ c[(x-3)-(y-2)+z]=0 \\ \text{or,}~~ x-y+z=1.

Hence, option (a) is correct.

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