Skip to content
Question

Q.Vectors a⃗=3i^−2j^+2k^\vec{a} = 3\hat{i} - 2\hat{j} + 2\hat{k} and b⃗=i^+2k^\vec{b} = \hat{i} + 2\hat{k} represent the adjacent sides of a parallelogram. Find the vectors representing the diagonals of the parallelogram. Also, find their lengths.

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The diagonals of a parallelogram are given by d⃗1=a⃗+b⃗\vec{d}_1 = \vec{a} + \vec{b} and d⃗2=a⃗−b⃗\vec{d}_2 = \vec{a} - \vec{b} (or b⃗−a⃗\vec{b} - \vec{a}). For a⃗=3i^−2j^+2k^\vec{a} = 3\hat{i} - 2\hat{j} + 2\hat{k} and b⃗=i^+2k^\vec{b} = \hat{i} + 2\hat{k}, the diagonals are 4i^−2j^+4k^4\hat{i} - 2\hat{j} + 4\hat{k} and 2i^−2j^2\hat{i} - 2\hat{j}, with lengths 66 and 222\sqrt{2} respectively.


Why the diagonal vectors are a⃗+b⃗\vec{a} + \vec{b} and a⃗−b⃗\vec{a} - \vec{b}

Imagine a parallelogram with adjacent sides a⃗\vec{a} and b⃗\vec{b} meeting at a common vertex. The diagonal that goes from that common vertex to the opposite vertex is simply the vector sum a⃗+b⃗\vec{a} + \vec{b} — you travel along a⃗\vec{a} then along b⃗\vec{b}, or vice versa.

The other diagonal connects the tips of a⃗\vec{a} and b⃗\vec{b} (when placed tail-to-tail). To go from the tip of a⃗\vec{a} to the tip of b⃗\vec{b}, you go backwards along a⃗\vec{a} and then forward along b⃗\vec{b}, giving b⃗−a⃗\vec{b} - \vec{a}. Equivalently, starting from the tip of b⃗\vec{b} to the tip of a⃗\vec{a} gives a⃗−b⃗\vec{a} - \vec{b}. Both are valid diagonal vectors — they differ only in direction, and their lengths are the same.

For a parallelogram with adjacent sides a⃗\vec{a} and b⃗\vec{b}:

Diagonal 1: d⃗1=a⃗+b⃗\text{Diagonal 1: } \vec{d}_1 = \vec{a} + \vec{b}

Diagonal 2: d⃗2=a⃗−b⃗(or b⃗−a⃗)\text{Diagonal 2: } \vec{d}_2 = \vec{a} - \vec{b} \quad (\text{or } \vec{b} - \vec{a})


Step-by-step computation

1. Find d⃗1=a⃗+b⃗\vec{d}_1 = \vec{a} + \vec{b}

Add component-wise:

  • i^\hat{i}: 3+1=43 + 1 = 4
  • j^\hat{j}: −2+0=−2-2 + 0 = -2
  • k^\hat{k}: 2+2=42 + 2 = 4

So d⃗1=4i^−2j^+4k^\vec{d}_1 = 4\hat{i} - 2\hat{j} + 4\hat{k}.

2. Find d⃗2=a⃗−b⃗\vec{d}_2 = \vec{a} - \vec{b}

Subtract component-wise:

  • i^\hat{i}: 3−1=23 - 1 = 2
  • j^\hat{j}: −2−0=−2-2 - 0 = -2
  • k^\hat{k}: 2−2=02 - 2 = 0

So d⃗2=2i^−2j^+0k^=2i^−2j^\vec{d}_2 = 2\hat{i} - 2\hat{j} + 0\hat{k} = 2\hat{i} - 2\hat{j}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.