Question

Suppose P is the point on the line joining  \((-9,4,5)\) and \((11,0,-1)\) that lies closest to the origin O. Then \(|OP|^2\) equals to?

• A. \(3\)
• B. \(4\)
• C. \(2\)
• D. \(9\)
• E. \(1\)

Answer 1

The Correct Option is D

To find the point on the line closest to the origin, we can use the concept of projection as mentioned before. The point on the line closest to the origin will be the point where the line intersects the line passing through the origin and perpendicular to the given line.

Let's first find the direction vector of the line passing through \((-9, 4, 5) \) and\( (11, 0, -1):\)

Direction vector \(= (11, 0, -1) - (-9, 4, 5) \)

\(= (11 + 9, 0 - 4, -1 - 5) \)

\(= (20, -4, -6)\)

Now, let's find the vector from the origin \((0, 0, 0)\) to any point \((x, y, z)\) on the line:

Vector \(OP = (x, y, z).\)

To find the point P  \(OP = (x, y, z)\) on the line closest to the origin, the vector OP should be perpendicular to the direction vector of the line. Hence, their dot product should be zero:

(OP) · (Direction vector) = \(0\)

\((x, y, z) · (20, -4, -6) = 0\)

\(20x - 4y - 6z = 0.\)

We need to find the value of \((x, y, z)\) that satisfy this equation and lie on the line passing through \((-9, 4, 5)\).

 We can use the parametric equation for the line:

\(x = -9 + 20t, y = 4 - 4t, z = 5 - 6t.\)

Substitute these expressions into the equation

\(20x - 4y - 6z = 0\)

\(20(-9 + 20t) - 4(4 - 4t) - 6(5 - 6t) = 0.\)

Solve for t:

\(-180 + 400t - 16 + 16t - 30 + 36t = 0\)

\(⇒-180 + 400t - 16 + 16t - 30 + 36t = 0\)

\(⇒400t + 16t + 36t = 180 + 16 + 30\)

\(⇒452t = 226\)

\(⇒t = \dfrac{226}{452}\)

\(⇒t = \dfrac{1}{2}\)

Now that we have the value of t, we can find the coordinates of point P:

\(x = -9 + 20(\dfrac{1}{2}) = -9 + 10 = 1, \)

\(y = 4 - 4(\dfrac{1}{2}) = 4 - 2 = 2\)

\(z = 5 - 6(\dfrac{1}{2}) = 5 - 3 = 2\)

So, the coordinates of point P are \((1, 2, 2)\).

Now, to find the distance squared (\(|OP|^2\)) from the origin to point P, we use the distance formula:

\(|OP|^{2} = x^2 + y^2 + z^2\).

\(|OP|^{2} = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9\)

Therefore, \(|OP|^{2}\) =\(9\)    (_Ans).

Answer 2

To find the point \( P \) on the line joining \( A(-9, 4, 5) \) and \( B(11, 0, -1) \) that lies closest to the origin \( O(0, 0, 0) \), we proceed as follows:

Step 1: Parametric Equation of the Line

The direction vector \( \vec{AB} \) is:

\[ \vec{AB} = B - A = (11 - (-9), 0 - 4, -1 - 5) = (20, -4, -6) \]

The parametric equations of the line can be written as:

\[ \vec{r}(t) = A + t \cdot \vec{AB} = (-9 + 20t, 4 - 4t, 5 - 6t) \]

Step 2: Distance Squared from Origin

The distance squared from the origin to a point \( P(t) \) on the line is:

\[ OP^2 = (-9 + 20t)^2 + (4 - 4t)^2 + (5 - 6t)^2 \]

Expanding this:

\[ OP^2 = 81 - 360t + 400t^2 + 16 - 32t + 16t^2 + 25 - 60t + 36t^2 \]

Combine like terms:

\[ OP^2 = (400t^2 + 16t^2 + 36t^2) + (-360t - 32t - 60t) + (81 + 16 + 25) \]

\[ OP^2 = 452t^2 - 452t + 122 \]

Step 3: Minimize the Distance

To find the minimum distance, we take the derivative of \( OP^2 \) with respect to \( t \) and set it to zero:

\[ \frac{d}{dt}(OP^2) = 904t - 452 = 0 \]

Solving for \( t \):

\[ 904t = 452 \implies t = \frac{452}{904} = \frac{1}{2} \]

Step 4: Compute \( OP^2 \) at \( t = \frac{1}{2} \)

Substitute \( t = \frac{1}{2} \) back into the expression for \( OP^2 \):

\[ OP^2 = 452\left(\frac{1}{2}\right)^2 - 452\left(\frac{1}{2}\right) + 122 \]

\[ OP^2 = 452 \cdot \frac{1}{4} - 226 + 122 \]

\[ OP^2 = 113 - 226 + 122 = 9 \]

Alternative Method: Projection

The vector from \( A \) to \( O \) is \( \vec{AO} = (9, -4, -5) \). The projection of \( \vec{AO} \) onto \( \vec{AB} \) gives the parameter \( t \):

\[ t = \frac{\vec{AO} \cdot \vec{AB}}{\vec{AB} \cdot \vec{AB}} = \frac{(9)(20) + (-4)(-4) + (-5)(-6)}{20^2 + (-4)^2 + (-6)^2} = \frac{180 + 16 + 30}{400 + 16 + 36} = \frac{226}{452} = \frac{1}{2} \]

Substituting \( t = \frac{1}{2} \) into the line equation gives \( P \):

\[ P = \left(-9 + 20 \cdot \frac{1}{2}, 4 - 4 \cdot \frac{1}{2}, 5 - 6 \cdot \frac{1}{2}\right) = (1, 2, 2) \]

The distance squared from \( O \) to \( P \) is:

\[ OP^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 \]