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Vector Algebra question

2022 · 29 Jun · Shift 1 · Q23
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Vector Algebra question

2022 · 29 Jun · Shift 1 · Q23

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→=αi^+3j^−k^\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat ka=αi+3j​−k, b→=3i^−βj^+4k^\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat kb=3i−βj​+4k and c→=i^+2j^−2k^\overrightarrow c = \widehat i + 2\widehat j - 2\widehat kc=i+2j​−2k where α, β∈R\alpha ,\,\beta \in Rα,β∈R, be three vectors. If the projection of a→\overrightarrow aa on c→\overrightarrow cc is 103{{10} \over 3}310​ and b→×c→=−6i^+10j^+7k^\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat kb×c=−6i+10j​+7k, then the value of α+β\alpha + \betaα+β is equal to :
  1. A
    3
  2. B
    4
  3. C
    5
  4. D
    6
View written solutionFree

Correct answer: A

  1. Given vectors

a⃗=αi^+3j^−k^,\vec a = \alpha \hat i + 3\hat j - \hat k,a=αi^+3j^​−k^, b⃗=3i^−βj^+4k^,\vec b = 3\hat i - \beta \hat j + 4\hat k,b=3i^−βj^​+4k^, c⃗=i^+2j^−2k^.\vec c = \hat i + 2\hat j - 2\hat k.c=i^+2j^​−2k^.

We need to find α+β\alpha + \betaα+β.

  1. Use the projection condition to find α\alphaα

The scalar projection of a⃗\vec aa on c⃗\vec cc is

projc⃗(a⃗)=a⃗⋅c⃗∣c⃗∣=103.\text{proj}_{\vec c}(\vec a) = \frac{\vec a \cdot \vec c}{|\vec c|} = \frac{10}{3}.projc​(a)=∣c∣a⋅c​=310​.

First compute:

a⃗⋅c⃗=α(1)+3(2)+(−1)(−2)=α+6+2=α+8.\vec a \cdot \vec c = \alpha(1) + 3(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8.a⋅c=α(1)+3(2)+(−1)(−2)=α+6+2=α+8.

Also,

∣c⃗∣=12+22+(−2)2=1+4+4=3.|\vec c| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = 3.∣c∣=12+22+(−2)2​=1+4+4​=3.

So,

α+83=103.\frac{\alpha+8}{3} = \frac{10}{3}.3α+8​=310​.

Hence,

α+8=10  ⟹  α=2.\alpha + 8 = 10 \implies \alpha = 2.α+8=10⟹α=2.

  1. Use the cross product condition to find β\betaβ

Given

b⃗×c⃗=−6i^+10j^+7k^.\vec b \times \vec c = -6\hat i + 10\hat j + 7\hat k.b×c=−6i^+10j^​+7k^.

Now,

b⃗=(3,−β,4),c⃗=(1,2,−2).\vec b = (3,-\beta,4), \quad \vec c = (1,2,-2).b=(3,−β,4),c=(1,2,−2).

Compute b⃗×c⃗\vec b \times \vec cb×c:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -\beta & 4\\ 1 & 2 & -2 \end{vmatrix}.$$ Expanding, $$\vec b \times \vec c = \hat i\big(( -\beta)(-2)-4\cdot 2\big) - \hat j\big(3(-2)-4(1)\big) + \hat k\big(3\cdot 2-(-\beta)(1)\big).$$ So, $$\vec b \times \vec c = \hat i(2\beta-8) - \hat j(-6-4) + \hat k(6+\beta).$$ Thus, $$\vec b \times \vec c = (2\beta-8)\hat i + 10\hat j + (\beta+6)\hat k.$$ Compare with $$-6\hat i + 10\hat j + 7\hat k.$$ Matching components: $$2\beta - 8 = -6 \implies 2\beta = 2 \implies \beta = 1.$$ Also, $$\beta + 6 = 7 \implies \beta = 1,$$ which is consistent. --- 4. **Find $\alpha + \beta$** $$\alpha + \beta = 2 + 1 = 3.$$ --- 5. **Option check** - A: $3$ ✅ - B: $4$ - C: $5$ - D: $6$ So the correct option is **A**.
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