Plane | Part-1 | Ex-5A Examhoop

Choose the correct option :

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

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

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