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CUET Mathematics 2025 PYQs15 May Shift 2. Free, no login required.

Algebra

past year questions, CUET Mathematics 2025 15 May Shift 2

Vector Algebra

past year questions, Algebra, CUET Mathematics 2025 15 May Shift 2

Medium

core

If a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} are three unit vectors such that a⃗+2b⃗−3c⃗=0⃗\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0}, then the value of 2a⃗.b⃗−6b⃗.c⃗−3c⃗.a⃗2\vec{a}.\vec{b} - 6\vec{b}.\vec{c} - 3\vec{c}.\vec{a} is

Solution

✅ Correct Option: 1

Since a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are unit vectors:

∣a⃗∣2=∣b⃗∣2=∣c⃗∣2=1|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1

The given equation a⃗+2b⃗−3c⃗=0⃗\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0} can be rearranged in three ways and squared to find each dot product.

a⃗+2b⃗=3c⃗\vec{a} + 2\vec{b} = 3\vec{c}

Taking the dot product of each side with itself:

∣a⃗∣2+4(a⃗⋅b⃗)+4∣b⃗∣2=9∣c⃗∣2|\vec{a}|^2 + 4(\vec{a}\cdot\vec{b}) + 4|\vec{b}|^2 = 9|\vec{c}|^2

1+4(a⃗⋅b⃗)+4=91 + 4(\vec{a}\cdot\vec{b}) + 4 = 9

a⃗⋅b⃗=1⋯(i)\vec{a}\cdot\vec{b} = 1 \quad \cdots (i)

a⃗=3c⃗−2b⃗\vec{a} = 3\vec{c} - 2\vec{b}

Taking the dot product of each side with itself:

∣a⃗∣2=9∣c⃗∣2−12(b⃗⋅c⃗)+4∣b⃗∣2|\vec{a}|^2 = 9|\vec{c}|^2 - 12(\vec{b}\cdot\vec{c}) + 4|\vec{b}|^2

1=9−12(b⃗⋅c⃗)+41 = 9 - 12(\vec{b}\cdot\vec{c}) + 4

12(b⃗⋅c⃗)=1212(\vec{b}\cdot\vec{c}) = 12

b⃗⋅c⃗=1⋯(ii)\vec{b}\cdot\vec{c} = 1 \quad \cdots (ii)

2b⃗=3c⃗−a⃗2\vec{b} = 3\vec{c} - \vec{a}

Taking the dot product of each side with itself:

4∣b⃗∣2=9∣c⃗∣2−6(c⃗⋅a⃗)+∣a⃗∣24|\vec{b}|^2 = 9|\vec{c}|^2 - 6(\vec{c}\cdot\vec{a}) + |\vec{a}|^2

4=9−6(c⃗⋅a⃗)+14 = 9 - 6(\vec{c}\cdot\vec{a}) + 1

6(c⃗⋅a⃗)=66(\vec{c}\cdot\vec{a}) = 6

c⃗⋅a⃗=1⋯(iii)\vec{c}\cdot\vec{a} = 1 \quad \cdots (iii)

Substituting (i)(i), (ii)(ii), and (iii)(iii):

2a⃗⋅b⃗−6b⃗⋅c⃗−3c⃗⋅a⃗2\vec{a}\cdot\vec{b} - 6\vec{b}\cdot\vec{c} - 3\vec{c}\cdot\vec{a}

=2(1)−6(1)−3(1)= 2(1) - 6(1) - 3(1)

=2−6−3= 2 - 6 - 3

=−7= \boxed{-7}

Related questions:

2025: 27 May Shift 1

Match List-I with List-II

List-IList-II
(A) If vector a⃗\vec{a} and b⃗\vec{b} are such that a⃗=λb⃗\vec{a} = \lambda \vec{b} and ∣a⃗∣=∣b⃗∣\vert \vec{a}\vert = \vert \vec{b}\vert , then(I) a⃗\vec{a} and b⃗\vec{b} are orthogonal
(B) Projection vector of a⃗\vec{a} on b⃗\vec{b}(II) [0,12][0, 12]
(C) a⃗\vec{a} and b⃗\vec{b} are non-zero vectors such that ∣a⃗+b⃗∣=∣a⃗−b⃗∣\vert \vec{a} + \vec{b}\vert = \vert \vec{a} - \vec{b}\vert , then(III) a⃗=±b⃗\vec{a} = \pm \vec{b}
(D) If ∣a⃗∣=4,−3≤λ≤2\vert \vec{a}\vert = 4, -3 \le \lambda \le 2, then the range of ∣λa⃗∣\vert \lambda \vec{a}\vert (IV) (a⃗⋅b⃗∣b⃗∣2)b⃗(\dfrac{\vec{a} \cdot \vec{b}}{\vert \vec{b}\vert ^2}) \vec{b}

Choose the correct answer from the options given below:

2026: 21 May Shift 1

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are unit vectors such that a⃗+b⃗+c⃗=0\vec{a} + \vec{b} + \vec{c} = 0. The value of a⃗.b⃗+b⃗.c⃗+c⃗.a⃗\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} is

2026: 29 May Shift 2

The value of λ\lambda for which the vectors 3i^+6j^−k^3\hat{i}+6\hat{j}-\hat{k} and −2i^+4j^−λk^-2\hat{i}+4\hat{j}-\lambda\hat{k} are perpendicular is

2026: 29 May Shift 2

If a⃗=2i^+j^\vec{a}=2\hat{i}+\hat{j}, and b⃗=j^−k^\vec{b}=\hat{j}-\hat{k}, then which of the following statements is/are TRUE ?

A. ∣a⃗+b⃗∣=3|\vec{a}+\vec{b}|=3

B. Unit vector along a⃗+b⃗\vec{a}+\vec{b} is 2i^+2j^−k^2\hat{i}+2\hat{j}-\hat{k}

C. Direction ratio of a⃗+b⃗\vec{a}+\vec{b} are 2, 2, -1

D. unit vector perpendicular to a⃗\vec{a} and b⃗\vec{b} is −13i^+23j^+23k^-\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}

Choose the correct answer from the options given below:

2025: 2 June Shift 1

Which of the following statements are true?

(A) If r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, then x,y,zx, y, z are called direction ratios of r⃗\vec{r}.

(B) For any two vectors a⃗\vec{a} and b⃗\vec{b}, a⃗+b⃗=b⃗+a⃗\vec{a} + \vec{b} = \vec{b} + \vec{a}

(C) a⃗⊥b⃗\vec{a} \perp \vec{b} if and only if a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}

(D) Projection of b⃗\vec{b} on a⃗\vec{a} is a⃗⋅b⃗∣a⃗∣2\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}

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2022: 6 Aug Shift 2

Let OA⃗=2i^−j^+k^\vec{OA} = 2\hat{i} - \hat{j} + \hat{k} and OB⃗=i^+j^−k^\vec{OB} = \hat{i} + \hat{j} - \hat{k}. Then

A. The magnitude of vector OA⃗\vec{OA} is 6

B. The magnitude of vector OB⃗\vec{OB} is 3\sqrt{3}

C. The vector AB⃗\vec{AB} is (−i^+2j^−2k^)(-\hat{i} + 2\hat{j} - 2\hat{k})

D. OA⃗⋅OB⃗=0\vec{OA} \cdot \vec{OB} = 0

E. OA⃗∥OB⃗\vec{OA} \parallel \vec{OB}

Choose the correct answer from the options given below:

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