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Chapter 16: Problem 1542

Perpendicular distance from point \((1,3,4)\) to line \([(x-5) / 2]=[(y-6) /(-1)]=[(z+7) / 3]\) is (A) \([\sqrt{(1398) / 7]}\) (B) \([\sqrt{(1398) / 14}]\) (D) \([(1398) / 7]\)

Short Answer

Expert verified
Perpendicular distance = \(\boxed{[\sqrt{(1398) / 14}]}\, (Option \, B)\)

Step by step solution

01

Identify a Vector Parallel to the Line

Let's first find a vector that is parallel to the given line. To do this, note that the line can be written parametrically as: \[x = 5 + 2t,\] \[y = 6 - t,\] \[z = -7 + 3t.\] So, the directional vector of the line will be given by \(\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\).
02

Find a Vector Connecting the Given Point to a Point on the Line

Next, we will find a vector connecting the given point to a point on the line. We can do this by choosing a value for the parameter \(t\); we can choose \(t=0\) to make our calculations simpler. The point on the line for \(t=0\) is \((5, 6, -7)\), and the position vector for the given point \((1, 3, 4)\) is \(\begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix}\). So, the required vector connecting the points will be: \[\begin{bmatrix} 1-5 \\ 3-6 \\ 4-(-7) \end{bmatrix} = \begin{bmatrix} -4 \\ -3 \\ 11 \end{bmatrix}.\]
03

Find the Area of the Parallelogram

Now, we need to find the area of the parallelogram formed by the vectors \(\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\) and \(\begin{bmatrix} -4 \\ -3 \\ 11 \end{bmatrix}\). We can use the vector cross product for this, so let's find the cross product of these two vectors: \[\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} \times \begin{bmatrix} -4 \\ -3 \\ 11 \end{bmatrix} = \begin{bmatrix} (-1)(11) - (3)(-3) \\ (3)(-4) - (2)(11) \\ (2)(-3) - (-1)(-4) \end{bmatrix} = \begin{bmatrix} -8 \\ -26 \\ -2 \end{bmatrix}.\] The area of the parallelogram will be the magnitude of this vector: \[\sqrt{(-8)^2 + (-26)^2 + (-2)^2} = \sqrt{64 + 676 + 4} = \sqrt{744}.\]
04

Calculate the Perpendicular Distance

Finally, we can find the perpendicular distance by dividing the area of the parallelogram by the magnitude of the direction vector, which in this case is \(\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\): \[Distance = \frac{\sqrt{744}}{\sqrt{(2)^2 + (-1)^2 + (3)^2}} = \frac{\sqrt{744}}{\sqrt{14}}.\] Comparing our result with the given options, we can see that the answer that matches our solution is: \[Perpendicular \, distance = \boxed{[\sqrt{(1398) / 14}]}\, (Option \, B)]\]

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Most popular questions from this chapter

Direction cosine of line \([(4-x) / 7]=[(y+9) / 5]=[(3 z+8) / 2]\) is (A) \(-[21 / \sqrt{(670)}],[15 / \sqrt{(670)}],[2 / \sqrt{(670)}]\) (B) \([21 / \sqrt{(670)}],[15 / \sqrt{(670)}],[2 / \sqrt{(670)}]\) (C) \([21 / \sqrt{(670)}],[(-15) / \sqrt{(670)}],[2 / \sqrt{(670)}]\) (D) \([(-21) / \sqrt{(670)}],[(-15) / \sqrt{(670)}],[(-2) / \sqrt{(670)}]\)

For vectors \(\underline{a}, \underline{b}, c \underline{c}\) if each vector is perpendicular to the sum of remaining two vectors and \(|\underline{\mathrm{a}}|=3,|\underline{\mathrm{b}}|=4,|\underline{\mathrm{c}}|=5\), then \(|\underline{a}+\underline{b}+\underline{c}|=\) (A) \(2 \sqrt{2}\) (B) \(3 \sqrt{2}\) (C) \(4 \sqrt{2}\) (D) \(5 \sqrt{2}\)

Lines \([\\{(\mathrm{k}+3),-\mathrm{k}-1, \mathrm{k}+1\\} / \mathrm{k} \in \mathrm{R}\\},\\{(2 \mathrm{k}, 0,3 \mathrm{k}-3) / \mathrm{k} \in \mathrm{R}\\}\) are (A) parallel (B) Intersecting (C) coincident (D) skew

If y-intercept of plane \((x-y+z-1)+\lambda(x+y-z-1)=0\) is 3 unit then \(\lambda=\) (A) \(-2\) (B) 2 (C) \((\mathrm{D})-(1 / 2)\)

For \(\underline{x}=(a, 3,-2), y=(a,-a, 2)\), if \(\underline{x} \perp y\), then \(a=\) (A) 4,1 (B) \(4,-1\) (C) \(-4,-1\) (D) \(-4,1\)
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