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A symmetrical form of the line of inters...

A symmetrical form of the line of intersection of the planes `x=ay+b` and `z=cy+d` is

A

`(x-b)/a=(y-1)/0=(z-d)/c`

B

`(x-b)/a=(y-0)/1=(z-d)/c`

C

`(x-a)/b=(y-0)/1=(z-c)/d`

D

none of these

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The correct Answer is:
To find the symmetrical form of the line of intersection of the planes given by the equations \( x = ay + b \) and \( z = cy + d \), we can follow these steps: ### Step 1: Express \( y \) in terms of \( x \) From the first plane equation \( x = ay + b \), we can rearrange it to express \( y \): \[ y = \frac{x - b}{a} \] **Hint:** Rearranging equations to isolate a variable is a common technique in solving systems of equations. ### Step 2: Substitute \( y \) into the second plane equation Now, we substitute the expression for \( y \) into the second plane equation \( z = cy + d \): \[ z = c\left(\frac{x - b}{a}\right) + d \] **Hint:** Substituting known values into equations helps to eliminate variables and find relationships between the remaining variables. ### Step 3: Simplify the equation for \( z \) Now, simplify the equation: \[ z = \frac{c(x - b)}{a} + d \] \[ z = \frac{cx - cb}{a} + d \] To combine the terms, we can express \( d \) in terms of a common denominator: \[ z = \frac{cx - cb + ad}{a} \] **Hint:** When combining fractions, ensure that you have a common denominator to simplify correctly. ### Step 4: Rearranging the equation Now, rearranging gives us: \[ az = cx - cb + ad \] \[ az - cx + cb - ad = 0 \] **Hint:** Rearranging terms helps to see the relationship between the variables more clearly. ### Step 5: Write in symmetrical form Now, we can express the equations in a symmetrical form. We have: \[ \frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c} \] **Hint:** Symmetrical forms are useful for visualizing the relationships between multiple variables. ### Final Result Thus, the symmetrical form of the line of intersection of the given planes is: \[ \frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c} \] ### Summary of Steps 1. Isolate \( y \) from the first plane equation. 2. Substitute \( y \) into the second plane equation. 3. Simplify the equation for \( z \). 4. Rearrange the equation to highlight the relationships. 5. Write the final result in symmetrical form.

To find the symmetrical form of the line of intersection of the planes given by the equations \( x = ay + b \) and \( z = cy + d \), we can follow these steps: ### Step 1: Express \( y \) in terms of \( x \) From the first plane equation \( x = ay + b \), we can rearrange it to express \( y \): \[ y = \frac{x - b}{a} \] ...
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OBJECTIVE RD SHARMA ENGLISH-PLANE AND STRAIGHT LINE IN SPACE -Chapter Test
  1. A symmetrical form of the line of intersection of the planes x=ay+b an...

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  2. The length of the perpendicular from the origin to the plane passing t...

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  3. The value of lamda for which the lines (x-1)/1=(y-2)/(lamda)=(z+1)/(-1...

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  4. The angle between the lines (x+4)/(1) = (y-3)/(2) = (z+2)/(3) and (x)/...

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  5. The direction cosines of the line 6x-2=3y+1=2z-2 are

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  6. A line passes through two points A(2,-3,-1) and B(8,-1,2). The coordin...

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  7. The position vector of a point at a distance of 3sqrt(11) units from h...

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  8. The line joining the points 6veca-4vecb+4vecc, -4vecc and the line joi...

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  9. The image (or reflection) of the point (1,2-1) in the plane vecr.(3hat...

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  10. The equation of the plane through the line of intersection of the plan...

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  11. Angle between the line vecr=(2hati-hatj+hatk)+lamda(-hati+hatj+hatk) a...

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  12. The line through hati+3hatj+2hatkandbot"to the line "vecr=(hati+2hatj-...

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  13. The distance of the point having position vector -hat(i) + 2hat(j) + 6...

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  14. The position vector of the point in which the line joining the points ...

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  15. The two lines vecr=veca+veclamda(vecbxxvecc) and vecr=vecb+mu(veccxxve...

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  16. Lines vecr = veca(1) + lambda vecb and vecr = veca(2) + svecb will lie...

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  17. Equation of a line passing through (-1,2,-3) and perpendicular to the ...

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  18. Find the Vector and Cartesian equation of line passing through (1, -2,...

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  19. The distance between the planes given by vecr.(hati+2hatj-2hatk)+5=0...

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  20. Find shortest distance between the line vecr = (5hati + 7hatj + 3ha...

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  21. Find the shortest distance between the lines vecr=(hatii+2hatj+hatk)+l...

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