Projections of Points and Lines Questions and Answers

Answer: A

Correct answer: perpendicular to the line

The shortest distance from a point to a line is along the perpendicular from the point to it - the only direction in which nothing is wasted travelling along the line. In projection the difficulty is that the perpendicular appears at true length only in a view where the line appears as a point, which is why such a view is constructed for the purpose.

Answer: B

Correct answer: perpendicular to the line, after a view showing its true length

The point view requires looking along the line, so the auxiliary plane must be perpendicular to it. That direction can only be established once a view showing the line at true length exists, since the line must be seen undistorted before the perpendicular to it can be set out. Two successive auxiliaries are therefore needed.

Answer: C

Correct answer: 40 mm above and 25 mm below the reference line

Height above the horizontal plane appears in the elevation, measured above the reference line, and distance in front of the vertical plane appears in the plan, measured below it after the planes are opened out. So the elevation sits 40 mm above and the plan 25 mm below, on the same projector.

Answer: D

Correct answer: 60.6 mm

The plan foreshortens the line by the cosine of its inclination to the horizontal plane: \[ 70 \cos 30^\circ = 70 \times 0.866 = 60.6\ \mathrm{mm}. \] Using the sine instead gives 35 mm, which is the height the line rises rather than the length it covers on plan.

Answer: A

Correct answer: 35 mm

The difference in height is the vertical component of the line: \[ 70 \sin 30^\circ = 70 \times 0.5 = 35\ \mathrm{mm}. \] The plan length and the height difference are the two perpendicular sides of the right triangle whose hypotenuse is the true length, which is the whole basis of the trapezoidal construction.

Answer: B

Correct answer: 75 mm

The plan length and the height difference are perpendicular, so the true length is the hypotenuse: \[ \sqrt{60^2 + 45^2} = \sqrt{3600 + 2025} = \sqrt{5625} = 75\ \mathrm{mm}. \] The 3-4-5 triangle appears again, scaled by fifteen, which is worth recognising since it saves the arithmetic.

Answer: C

Correct answer: 36.9 degrees

The true inclination is the angle whose tangent is the height difference over the plan length: \[ \tan\theta = \frac{45}{60} = 0.75, \qquad \theta = 36.9^\circ. \] Note that this is less than the apparent inclination measured in the elevation, as it must be - the projection always overstates the angle.

Answer: D

Fifty and sixty sum to a hundred and ten, which no line can achieve, since the two inclinations are geometrically constrained to ninety degrees together. A pair such as thirty and forty-five would be perfectly admissible.

Answer: A

Correct answer: 50 degrees

For any line the two true inclinations satisfy \( \theta + \phi \le 90^\circ \), the sum reaching ninety only when the line lies in a profile plane. With 40 degrees to the horizontal plane, the greatest possible inclination to the vertical plane is \( 90 - 40 = 50 \) degrees. A pair summing to more than ninety describes no line at all.

Answer: B

Correct answer: a point

A vertical line is perpendicular to the horizontal plane, so it is seen end on from above and its plan is a single point where both ends coincide. Its length appears in full in the elevation, since the line is parallel to the vertical plane, so nothing is lost - the information is simply divided between the two views.