Board 2024 Class 12 Maths Q15 Solution: Vector from B(3, −4, 7) to A(2, −3, 5)

The question

The vector with terminal point $A(2, -3, 5)$ and initial point $B(3, -4, 7)$ is :

(A) $\hat i - \hat j + 2\hat k$

(B) $\hat i + \hat j + 2\hat k$

(C) $-\hat i - \hat j - 2\hat k$

(D) $-\hat i + \hat j - 2\hat k$

Exam answer

The vector goes from the initial point $B$ to the terminal point $A$ ⁠:

\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB}

Why?

The vector joining two points is the position vector of the terminal point minus that of the initial point:

\begin{aligned} &\overrightarrow{PQ} = (x_2 - x_1)\hat i \\ &\ + (y_2 - y_1)\hat j + (z_2 - z_1)\hat k \end{aligned}

for initial point $P(x_1, y_1, z_1)$ and terminal point $Q(x_2, y_2, z_2)$ ⁠.

\begin{aligned} \overrightarrow{BA} &= (2 - 3)\hat i + (-3 + 4)\hat j \\ &\quad + (5 - 7)\hat k \\ &= -\hat i + \hat j - 2\hat k \end{aligned}

Why?

Each component is a coordinate of $A(2, -3, 5)$ minus the matching coordinate of $B(3, -4, 7)$ ⁠: $2 - 3 = -1$ ⁠, $-3 - (-4) = 1$ and $5 - 7 = -2$ ⁠.

\begin{aligned} x&: 2 - 3 = -1 \\ y&: -3 - (-4) \\ &\ \ = -3 + 4 = 1 \\ z&: 5 - 7 = -2 \end{aligned}

Watch out: subtracting the other way gives $\overrightarrow{AB} = \hat i - \hat j + 2\hat k$ ⁠, which is option (A), the vector pointing the opposite way.

Answer: (D) $-\hat i + \hat j - 2\hat k$

Where this comes from

Chapter 10, Vector Algebra: Vector joining two points.

Question type: The vector from A to B, and its direction cosines

How to spot it: click to see

Type 2. The vector from A to B, and its direction cosines

Spot it: two points are given, like $A(1, 2, -3)$ and $B(-1, -2, 1)$ ⁠.

Here: The start is $B(3, -4, 7)$ and the end is $A(2, -3, 5)$ ⁠. Take A minus B part by part to get $-\hat i + \hat j - 2\hat k$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Q15: Vector from B(3,−4,7)B(3, -4, 7)B(3,−4,7) to A(2,−3,5)A(2, -3, 5)A(2,−3,5)

Q15: Vector from $B(3, -4, 7)$ to $A(2, -3, 5)$