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Q.The vector with terminal point A (2,−3,5)(2,-3,5) and initial point B (3,−4,7)(3,-4,7) is :
(A) i^−j^+2k^\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}
(B) i^+j^+2k^\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}
(C) −i^−j^−2k^-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}
(D) −i^+j^−2k^-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★
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The vector from initial point B to terminal point A is found by subtracting the coordinates of B from A. The result is −i^+j^−2k^-\hat{i} + \hat{j} - 2\hat{k}, which matches option (D).

The key idea here is simple but often flipped: a vector is defined by its terminal minus initial coordinates. If you mix up which point is which, you'll get the opposite sign — a classic trap in vector problems.

Let’s break it down.

  1. Understand what the question asks We are given terminal point A (2,−3,5)(2, -3, 5) and initial point B (3,−4,7)(3, -4, 7). The vector from B to A is written as BA→\overrightarrow{BA} (or sometimes v⃗\vec{v} with tail at B and head at A). The formula is:

BA→=(coordinates of terminal point)−(coordinates of initial point)\overrightarrow{BA} = \text{(coordinates of terminal point)} - \text{(coordinates of initial point)}

  1. Apply the formula component-wise

    Subtract B’s coordinates from A’s:

    • xx-component: 2−3=−12 - 3 = -1
    • yy-component: −3−(−4)=−3+4=1-3 - (-4) = -3 + 4 = 1
    • zz-component: 5−7=−25 - 7 = -2
  2. Write the vector in unit vector form

    The components (−1,1,−2)(-1, 1, -2) correspond to:

−1i^+1j^−2k^=−i^+j^−2k^-1\hat{i} + 1\hat{j} - 2\hat{k} = -\hat{i} + \hat{j} - 2\hat{k}

  1. Match with the options Looking at the choices:
    • (A) i^−j^+2k^\hat{i} - \hat{j} + 2\hat{k}
    • (B) i^+j^+2k^\hat{i} + \hat{j} + 2\hat{k}
    • (C) −i^−j^−2k^-\hat{i} - \hat{j} - 2\hat{k} …

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