If a vector 2 i +3 j +8 k is perpendicular to the vector 4 j -4 i +α k , then the value of α is

Correct answer: C. -(1)/(2) Two vectors must be perpendicular if their dot product is zero. Let, aligned a & =2 i +3 j +8 k b & =4 j -4 i +α k & =-4 i +4 j +α k aligned According to the above hypothesis a b a · b =0 array lc & (2 i +3 j +8 k ) ·(-4 i +4 j +α k )=0 & -8+12+8 α=0 & 8 α=-4 & α=-(4)/(8)=-(1)/(2) array NOTE: a · b = a b θ. Here, a and b are always positive as they are the magnitudes a and b .

JIPMER 2007

Exam: JIPMER
Year: 2007
Subject: Physics
Chapter: Mathematics in Physics

JIPMER 2007 · Physics · Mathematics in Physics

If a vector 2 i +3 j +8 k is perpendicular to the vector 4 j -4 i...

If a vector

2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}

is perpendicular to the vector

4 \hat{\mathbf{j}}-4 \hat{\mathbf{i}}+\alpha \hat{\mathbf{k}}

, then the value of

\alpha

Options

A $-1$
B $\frac{1}{2}$
C $-\frac{1}{2}$
D $1$

Correct answer

C. $-\frac{1}{2}$

Step-by-step solution

Two vectors must be perpendicular if their dot product is zero. Let,

\begin{aligned} \overrightarrow{\mathbf{a}} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} \\ \overrightarrow{\mathbf{b}} & =4 \hat{\mathbf{j}}-4 \hat{\mathbf{i}}+\alpha \hat{\mathbf{k}} \\ & =-4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}\end{aligned}

According to the above hypothesis

\overrightarrow{\mathbf{a}} \perp \overrightarrow{\mathbf{b}} \Rightarrow\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0

\begin{array}{lc}\Rightarrow & (2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \cdot(-4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})=0 \\ \Rightarrow & -8+12+8 \alpha=0 \\ \Rightarrow & 8 \alpha=-4 \\ \therefore & \alpha=-\frac{4}{8}=-\frac{1}{2}\end{array}

NOTE:

\overrightarrow{

\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=

a b

\cos \theta. Here,

a and b

are always positive as they are the magnitudes

\overrightarrow{

\mathbf{a}}

and

\overrightarrow{\mathbf{b}

}

.

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