Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Vector Algebra question

2020 · 7 Jan · Shift 1 · Q26
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Vector Algebra
  5. /2020 · 7 Jan · Shift 1 · Q26

Vector Algebra question

2020 · 7 Jan · Shift 1 · Q26

JEE MainMathematicsVector AlgebraMCQ+4 / −1
A vector a→=αi^+2j^+βk^(α,β∈R)\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)a=αi+2j​+βk(α,β∈R) lies in the plane of the vectors, b→=i^+j^\overrightarrow b = \widehat i + \widehat jb=i+j​ and c→=i^−j^+4k^\overrightarrow c = \widehat i - \widehat j + 4\widehat kc=i−j​+4k. If a→\overrightarrow aa bisects the angle between b→\overrightarrow bb and c→\overrightarrow cc, then:
  1. A
    a→.i^+3=0\overrightarrow a .\widehat i + 3 = 0a.i+3=0
  2. B
    a→.k^−4=0\overrightarrow a .\widehat k - 4 = 0a.k−4=0
  3. C
    a→.i^+1=0\overrightarrow a .\widehat i + 1 = 0a.i+1=0
  4. D
    a→.k^+2=0\overrightarrow a .\widehat k + 2 = 0a.k+2=0
View written solutionFree

Correct answer: B

  1. Given vectors

a⃗=αi^+2j^+βk^,\vec a=\alpha \hat i+2\hat j+\beta \hat k,a=αi^+2j^​+βk^, b⃗=i^+j^=(1,1,0),\vec b=\hat i+\hat j=(1,1,0),b=i^+j^​=(1,1,0), c⃗=i^−j^+4k^=(1,−1,4).\vec c=\hat i-\hat j+4\hat k=(1,-1,4).c=i^−j^​+4k^=(1,−1,4).

We are told that:

  • a⃗\vec aa lies in the plane of b⃗\vec bb and c⃗\vec cc,
  • a⃗\vec aa bisects the angle between b⃗\vec bb and c⃗\vec cc.

We must determine which option is correct.

  1. Use the angle bisector direction formula

The internal angle bisector of vectors b⃗\vec bb and c⃗\vec cc has direction

b⃗∣b⃗∣+c⃗∣c⃗∣.\frac{\vec b}{|\vec b|}+\frac{\vec c}{|\vec c|}.∣b∣b​+∣c∣c​.

First compute magnitudes:

∣b⃗∣=12+12=2,|\vec b|=\sqrt{1^2+1^2}=\sqrt2,∣b∣=12+12​=2​, ∣c⃗∣=12+(−1)2+42=18=32.|\vec c|=\sqrt{1^2+(-1)^2+4^2}=\sqrt{18}=3\sqrt2.∣c∣=12+(−1)2+42​=18​=32​.

Hence,

b⃗∣b⃗∣=12(1,1,0),\frac{\vec b}{|\vec b|}=\frac{1}{\sqrt2}(1,1,0),∣b∣b​=2​1​(1,1,0), c⃗∣c⃗∣=132(1,−1,4).\frac{\vec c}{|\vec c|}=\frac{1}{3\sqrt2}(1,-1,4).∣c∣c​=32​1​(1,−1,4).

So the bisector direction is

12(1,1,0)+132(1,−1,4).\frac{1}{\sqrt2}(1,1,0)+\frac{1}{3\sqrt2}(1,-1,4).2​1​(1,1,0)+32​1​(1,−1,4).

Taking common factor 132\frac{1}{3\sqrt2}32​1​,

=\frac{1}{3\sqrt2}(4,2,4).$$ Thus the direction ratio is $$\vec a \parallel (4,2,4)=(2,1,2).$$ --- 3. **Match with the given form of** $\vec a$ Given $$\vec a=(\alpha,2,\beta).$$ Since $\vec a$ is parallel to $(2,1,2)$, let $$\vec a=\lambda(2,1,2).$$ Then comparing the $j$-component: $$\lambda\cdot 1=2 \implies \lambda=2.$$ Therefore, $$\vec a=2(2,1,2)=(4,2,4).$$ So, $$\alpha=4,\qquad \beta=4.$$ --- 4. **Check the options** Since $$\vec a\cdot \hat i=4, \qquad \vec a\cdot \hat k=4,$$ we test each statement: - **A:** $\vec a\cdot \hat i+3=4+3=7\neq 0$ → false - **B:** $\vec a\cdot \hat k-4=4-4=0$ → true - **C:** $\vec a\cdot \hat i+1=4+1=5\neq 0$ → false - **D:** $\vec a\cdot \hat k+2=4+2=6\neq 0$ → false --- 5. **Final answer** The correct option is $$\boxed{\text{B}}$$
PreviousNext

More from Vector Algebra

  • Let a, b and c be three unit vectors such that a+b+c=0. If λ=a.b+b.c+c.a…2020 · MCQ
  • Let a=i−2j​+k and b=i−j​+k be two vectors. If c is a vector such that b×c=b×a…2020 · MCQ
  • Let a, b and c be three vectors such that ​a​=3​, ​b​=5,b.c=10 and…2020 · Numerical
  • Let a→=3i∧​+2j∧​+xk∧​ and b→​=i∧​−j∧​+k∧​, for some real…2019 · MCQ
  • Let α=3i+j​ and β​=2i−j​+3k. If β​=β​1​−β2​​, where β​1​…2019 · MCQ
  • If a unit vector a makes angles π/3 with i, π/ 4 with j​ and θ∈(0, π) with k, then a value of θ is :-2019 · MCQ
  • Let a=i+j​+2​k,b=b1​i+b2​j​+2​k, c=5i+j​+2​k be three vectors…2019 · MCQ
  • Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos (∠ GOA) (O being the origin) is equal to :2019 · MCQ