if the origin is shifted 12 3 without changing the

If the origin is shifted $$\left( {1,\,2,\, - 3} \right)$$ without changing the directions of the axis, then find the new coordinates of the point $$\left( {0,\,4,\,5} \right)$$ with respect to new frame.

A. $$\left( { - 1,\,2,\,8} \right)$$

B. $$\left( {4,\,5,\,1} \right)$$

C. $$\left( {3,\, - 2,\,4} \right)$$

D. $$\left( {6,\,0,\,8} \right)$$

Answer : $$\left( { - 1,\,2,\,8} \right)$$

Solution :
In the new frame $$x' = x - {x_1},\,y' = y - {y_1},\,z' = z - {z_1},\,$$
Where $$\left( {{x_1},\,{y_1},\,{z_1}} \right)$$ is shifted origin.
$$ \Rightarrow x' = 0 - 1 = - 1,\,y' = 4 - 2 = 2,\,z' = 5 + 3 = 8$$
Hence, the coordinates of the point with respect to the new coordinates frame are $$\left( { - 1,\,2,\,8} \right).$$

Releted MCQ Question on
Geometry >> Three Dimensional Geometry

Releted Question 1

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :

A. $$7$$

B. $$ - 7$$

C. no real value

D. $$4$$

View Answer

Releted Question 2

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :

A. $$\frac{3}{2}$$

B. $$\frac{9}{2}$$

C. $$ - \frac{2}{9}$$

D. $$ - \frac{3}{2}$$

View Answer

Releted Question 3

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :

A. $$0$$

B. $$1$$

C. $$\sqrt 2 $$

D. $$2\sqrt 2 $$

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Releted Question 4

Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :

A. $$\frac{1}{4}$$

B. $$ - \frac{1}{4}$$

C. $$\frac{1}{8}$$

D. $$ - \frac{1}{8}$$

View Answer

If the origin is shifted $$\left( {1,\,2,\, - 3} \right)$$ without changing the directions of the axis, then find the new coordinates of the point $$\left( {0,\,4,\,5} \right)$$ with respect to new frame.

Releted MCQ Question on
Geometry >> Three Dimensional Geometry

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :

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Three Dimensional Geometry

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If the origin is shifted $$\left( {1,\,2,\, - 3} \right)$$ without changing the directions of the axis, then find the new coordinates of the point $$\left( {0,\,4,\,5} \right)$$ with respect to new frame.

Releted MCQ Question on Geometry >> Three Dimensional Geometry

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :

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Releted MCQ Question on
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Three Dimensional Geometry