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Thought-Process to Discover Knowledge

Welcome to **nub****trek**.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

Welcome to **nub****trek**.

The content is presented in small-focused learning units to enable you to

think,

figure-out, &

learn.

**Just keep tapping** (or clicking) on the content to continue in the trail and learn. continue

The content is presented in small-focused learning units to enable you to

think,

figure-out, &

learn.

To make best use of nubtrek, understand what is available.

nubtrek is designed to explain mathematics and science for young readers. Every topic consists of four sections.

nub,

trek,

jogger,

exercise.

This captures the small-core of concept in simple-plain English. The objective is to make the learner to think about. continue

Trekking is bit hard, requiring one to sweat and exert. The benefits of taking the steps are awesome. In the trek, concepts are explained with exploratory questions and your thinking process is honed step by step. continue

This captures the essence of learning and helps one to review at a later point. The reference is available in pdf document too. This is designed to be viewed in a smart-phone screen. continue

This part does not have much content as of now. Over time, when resources are available, this section will have curated and exam-prep focused questions to test your knowledge. continue

*summary of this topic*

Voice

Voice

Home

» multiplication of a vector by a scalar is distributive over vector addition

→ `lambda (vec p + vec q) = lambda vec p + lambda vec q`

*plain and simple summary*

nub

*plain and simple summary*

nub

dummy

• A scalar multiplied by sum of two vectors equals sum of vectors multiplied by the scalar.

*simple steps to build the foundation*

trek

*simple steps to build the foundation*

trek

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In this page, the property, multiplication of scalar is distributive over vector addition, is explained.

Starting on learning the property "Distributive over Vector Addition". ;; In this page, the property, multiplication of scalar is distributive over vector addition, is explained.

Considering that a vector has components as real numbers, Given that `vec p = ai+bj+ck`, `vec q = e i + fj + g k` and `lambda ` where `a,b,c,d,e,f,lambda in bbb R`, What will be the result of `lambda (vec p + vec q)`?

- scalar
- `lambda vec p + lambda vec q`
- `lambda vec p + vec q`

The answer is '`lambda vec p + lambda vec q`'. This is proven by multiplying the components individually by the scalar and using vector addition properties.

*comprehensive information for quick review*

Jogger

*comprehensive information for quick review*

Jogger

dummy

**Distributive over Vector Addition: ** Given scalar `lambda` and vectors `vec p, vec q`,

`lambda(vec p + vec q) = lambda vec p + lambda vec q`

*practice questions to master the knowledge*

Exercise

*practice questions to master the knowledge*

Exercise

Given that `vec p = 6 vec q` and `vec r = 4 vec s`, which of the following equals `3 ( 2vec q+ 4/3 vec s)`?

- `vec p + vec r`
- `vec p + 2 vec r`
- `vec p - vec r`
- `vec r - vec p`

The answer is '`vec p + vec r`'.

*your progress details*

Progress

*About you*

Progress

Considering that a vector has components as real numbers. Given that vector p = a i+b j+c k, vector q = e i + f j + g k and lambda where a,b,c,d,e,f,lambda are in real numbers, ;; What will be the result of lambda multiplied vector p + vector q?

1

2

3

The answer is "lambda multiplied by vec p + lambda multiplied by vec q ". This is proven by multiplying the components individually by the scalar and using vector addition properties.

A scalar multiplied by sum of two vectors equals sum of vectors multiplied by the scalar.

Distributive over Vector Addition: Given scalar lambda and vectors p, vector q, lambda multiplied (vector p + vector q) = lambda times vector p + lambda times vector q

Given that vector p = 6 times vector q and vector r = 4 times vector s, which of the following equals 3 times, 2 vector q + 4 by 3 vector s?

1

2

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The answer is "vector p plus vector r".