Let P(3, 2, 3), Q(4, 6, 2), R(7, 3, 2) be the vertices of PQR. Then, the angle QPR is:

Given: The vertices are P(3, 2, 3), Q(4, 6, 2) and R(7, 3, 2). Find: The angle QPR. Use the dot product formula for the angle between the sides through point P. First, calculate the vectors: QP = P - Q = (3 - 4, 2 - 6, 3 - 2) = (-1, -4, 1) RP = P - R = (3 - 7, 2 - 3, 3 - 2) = (-4, -1, 1) Now, calculate the dot product: QP RP = (-1)(-4) + (-4)(-1) + (1)(1) = 4 + 4 + 1 = 9 Find the magnitudes: QP = (-1)^2 + (-4)^2 + 1^2 = 1 + 16 + 1 = 18 RP = (-4)^2 + (-1)^2 + 1^2 = 16 + 1 + 1 = 18 Substitute into the cosine formula: = QP RP QP RP = 9 18 18 = 918 = 12 Therefore, = ^-1 ( 12) = 3 Therefore, the correct option is D. Given: The vertices are P(3, 2, 3), Q(4, 6, 2) and R(7, 3, 2). Find: The angle QPR. Take direction ratios of the sides through P. For PR and PQ: PR = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1) PQ = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1) Now compute: = 4 1 + 1 4 + (-1) (-1) 18 18 = 4 + 4 + 118 = 918 = 12 Hence, = ^-1 ( 12) = 3 So, the correct option is D.

Let P(3, 2, 3), Q(4, 6, 2), R(7, 3, 2) be the vertices of PQR. Th… — Solution | JEE Mathematics

MCQEasyJEE 2024Coordinates in 3D

Let $P(3, 2, 3)$ , $Q(4, 6, 2)$ , $R(7, 3, 2)$ be the vertices of $\triangle PQR$ . Then, the angle $\triangle QPR$ is:

A
$\pi/6$
B
$\cos^{-1}(7/18)$
C
$\cos^{-1}(1/18)$
D
$\pi/3$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The vertices are $P(3, 2, 3)$ , $Q(4, 6, 2)$ and $R(7, 3, 2)$ .

Find: The angle $\angle QPR$ .

Use the dot product formula for the angle between the sides through point $P$ .

First, calculate the vectors:

\overrightarrow{QP} = P - Q = (3 - 4, 2 - 6, 3 - 2) = (-1, -4, 1)

\overrightarrow{RP} = P - R = (3 - 7, 2 - 3, 3 - 2) = (-4, -1, 1)

Now, calculate the dot product:

\overrightarrow{QP} \cdot \overrightarrow{RP} = (-1)(-4) + (-4)(-1) + (1)(1) = 4 + 4 + 1 = 9

Find the magnitudes:

\left\|\overrightarrow{QP}\right\| = \sqrt{(-1)^2 + (-4)^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18}

\left\|\overrightarrow{RP}\right\| = \sqrt{(-4)^2 + (-1)^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18}

Substitute into the cosine formula:

\cos \theta = \frac{\overrightarrow{QP} \cdot \overrightarrow{RP}}{\left\|\overrightarrow{QP}\right\|\left\|\overrightarrow{RP}\right\|} = \frac{9}{\sqrt{18} \times \sqrt{18}} = \frac{9}{18} = \frac{1}{2}

Therefore,

\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}

Therefore, the correct option is D.

Using direction ratios

Given: The vertices are $P(3, 2, 3)$ , $Q(4, 6, 2)$ and $R(7, 3, 2)$ .

Find: The angle $\angle QPR$ .

Take direction ratios of the sides through $P$ .

For $PR$ and $PQ$ :

PR = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)

PQ = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)

Now compute:

\cos \theta = \frac{4 \cdot 1 + 1 \cdot 4 + (-1) \cdot (-1)}{\sqrt{18} \cdot \sqrt{18}} = \frac{4 + 4 + 1}{18} = \frac{9}{18} = \frac{1}{2}

Hence,

\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}

So, the correct option is D.

Common mistakes

Using the angle at the wrong vertex is a common mistake. $\angle QPR$ is the angle at $P$ , so the vectors must be taken along the sides through $P$ such as $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ . Do not use vectors based at $Q$ or $R$ for a different angle.
Some students subtract coordinates inconsistently while forming vectors. A sign error in $\overrightarrow{PQ}$ or $\overrightarrow{PR}$ changes the dot product. Always subtract coordinates component-wise in the same order.
Confusing the dot product formula with the distance formula leads to error. To find an angle, use $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$ , not only side lengths without relating them through the angle formula.

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