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Chapter 16: Problem 1547

\(\mathrm{P}(1,6,3)\) to \((\mathrm{x} / 1)=[(\mathrm{y}-1) / 2]=[(\mathrm{z}-2) / 3]\) on then image of \(\mathrm{P}\) is (A) \(\overline{(-1,0,-7)}\) (B) \((-1,0,7)\) (C) \((1,0,7)\) (D) \((1,0,-7)\)

Short Answer

Expert verified
The given problem statement is incorrect as the provided line equations and point P do not intersect. Please review the original problem statement to ensure the correct values are provided before attempting to solve again.

Step by step solution

01

Write the parametric form of the line

Given: \((\mathrm{x} / 1)=[(\mathrm{y}-1) / 2]=[(\mathrm{z}-2) / 3]\). Let the parameter be \(t\). Now we can rewrite the equation in the parametric form: \(\mathrm{x} = 1t\) \(\mathrm{y} = 1 + 2t\) \(\mathrm{z} = 2 + 3t\) Thus, the parametric form of the line is \((1t, 1+2t, 2+3t)\).
02

Set the line equations equal to the point P's coordinates

Given point P as \((1,6,3)\). Now set the line equations equal to P: \(1t = 1\) \(1 + 2t = 6\) \(2 + 3t = 3\)
03

Solve for the parameter t

Let's solve the first equation: \(1t = 1\) \(t = 1\) Now substitute this value of t into the second and third equations to check if it satisfies both: \(1 + 2 (1) = 3\) \(2 + 3 (1) = 5\) As the parameter t does not satisfy either of the remaining equations, there is an error in the problem statement. The line equations given do not intersect the point P (\((1, 6, 3)\)). It's possible that there may have been a typo in the given exercise. Review the original problem statement and ensure that the correct values are provided before attempting to solve again.

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Most popular questions from this chapter

The equation of the locus of point which are equidistance from \((4,5,2)\) and \((1,6,3)\) is (A) \(6 x-2 y-2 z+1=0\) (B) \(6 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}+1=0\) (C) \(6 \mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+1=0\) (D) \(6 \mathrm{x}-2 \mathrm{y}-2 \mathrm{z}-1=0\)

If angle between two unit vectors \(\underline{a} \& \underline{b}\) is \(\alpha\), then \(|\underline{\mathrm{a}}-\underline{\mathrm{b}} \cos \alpha|=0<\alpha<(\pi / 2)\) (A) \(\sin \alpha\) (B) \(\sin (\alpha / 2)\) (C) \(\sin 2 \alpha\) (D) \(\sin ^{2}(\alpha / 2)\)

The equation of the line of the intersection of the planes \(x+2 y-3 z=6\) and \(2 x-y+z=7\) is (A) \([(\mathrm{x}-4) / 1]=[(\mathrm{y}-1) / 7]=(\mathrm{z} / 5)\) (B) \([(\mathrm{x}+4) / 1]=[(\mathrm{y}-1) / 7)=(\mathrm{z} / 5)\) (C) \([(\mathrm{x}+1) / 1]=[(\mathrm{y}+1) / 7]=(\mathrm{z} / 5)\) (D) \([(\mathrm{x}-1) /(-1)]=[(\mathrm{y}-1) /(-7)]=(\mathrm{z} / 5)\)

If \(\mathrm{m} \angle \mathrm{B}=(\pi / 2)\) in \(\Delta \mathrm{ABC}\) and \(\mathrm{P}, \mathrm{Q}\) are points of trisection of hypotenuse \(\underline{\mathrm{A}} \mathrm{C}\), then \(\mathrm{BP}^{2}+\mathrm{BQ}^{2}=\) (A) \((5 / 9) \mathrm{AC}^{2}\) (B) \((5 / 9) \mathrm{AC}\) (C) \((25 / 81) \mathrm{AC}^{2}\) (D) \((25 / 81) \mathrm{AC}\)

Line \(\mathrm{x}=2 \mathrm{y}+1,2 \mathrm{y}=1-\mathrm{z}\) and \(2 \mathrm{x}+\mathrm{y}+\mathrm{z}=0, \mathrm{z}+2=0\) angle between two line (A) 0 (B) \(\overline{(\pi / 4)}\) (C) \((\pi / 3)\) (D) \((\pi / 2)\)
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