If the equation of the plane passing through the point (-1,2,0) and parallel to the lines (x/3)=(y+1/0)=(z-2/-1) and (x-1/1)=(y+1/2)=(z+1/-1) is a x+b y+c z=1, then the value of (a+b+c), is

Question

If the equation of the plane passing through the point (-1,2,0) and parallel to the lines (x/3)=(y+1/0)=(z-2/-1) and (x-1/1)=(y+1/2)=(z+1/-1) is a x+b y+c z=1, then the value of (a+b+c), is (A) 3 (B) 4 (C) 5 (D) 10

Tardigrade Admin · Accepted Answer

Normal vector of the plane h=| hati hatj hatk 3 0 -1 1 2 -1| h=2 hat i +2 hat j +6 hat k =2( hat i + hat j +3 hat k ) ∴ Equation of plane 1( x +1)+1( y -2)+3( z -0)=0 P: x + y +3 z =1 hence, (a+b+c)=1+1+3=5.

Q. If the equation of the plane passing through the point $(- 1, 2, 0)$ and parallel to the lines $\frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 1}$ is $a x + b y + cz = 1$ , then the value of $(a + b + c)$ , is

3

4

5

10

Normal vector of the plane $h = ∣ ∣ \hat{i} 31 \hat{j} 02 \hat{k} - 1 - 1 ∣ ∣$
$h = 2 \hat{i} + 2 \hat{j} + 6 \hat{k} = 2 (\hat{i} + \hat{j} + 3 \hat{k})$
$∴$ Equation of plane $1 (x + 1) + 1 (y - 2) + 3 (z - 0) = 0$
P: $x + y + 3 z = 1$
hence, $(a + b + c) = 1 + 1 + 3 = 5$ .

Q. If the equation of the plane passing through the point (−1,2,0) and parallel to the lines 3x​=0y+1​=−1z−2​ and 1x−1​=2y+1​=−1z+1​ is ax+by+cz=1, then the value of (a+b+c), is

3

4

5

10

Normal vector of the plane h=∣∣​i^31​j^​02​k^−1−1​∣∣​ h=2i^+2j^​+6k^=2(i^+j^​+3k^) ∴ Equation of plane 1(x+1)+1(y−2)+3(z−0)=0 P: x+y+3z=1 hence, (a+b+c)=1+1+3=5.

Q. If the equation of the plane passing through the point $(- 1, 2, 0)$ and parallel to the lines $\frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 1}$ is $a x + b y + cz = 1$ , then the value of $(a + b + c)$ , is

Normal vector of the plane $h = ∣ ∣ \hat{i} 31 \hat{j} 02 \hat{k} - 1 - 1 ∣ ∣$
$h = 2 \hat{i} + 2 \hat{j} + 6 \hat{k} = 2 (\hat{i} + \hat{j} + 3 \hat{k})$
$∴$ Equation of plane $1 (x + 1) + 1 (y - 2) + 3 (z - 0) = 0$
P: $x + y + 3 z = 1$
hence, $(a + b + c) = 1 + 1 + 3 = 5$ .